Signals and Systems — Chapter 2: Fourier Series

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2069 Bhadra · 8 marks

Derive the exponential form of the Fourier series of a continuous-time periodic signal x(t) (synthesis and analysis equations).

Answer

A continuous-time periodic signal $x(t)$ with fundamental period $T$ and fundamental frequency $\omega_0 = 2\pi/T$ can be written as a weighted sum of harmonically related complex exponentials. This is the exponential Fourier series.

Synthesis equation

The set $\phi_k(t) = e^{jk\omega_0 t}$ , $k = 0, \pm1, \pm2, \dots$ , contains signals that are all periodic with period $T$ (the $k$ th one has frequency $k\omega_0$ ). A linear combination of them is also periodic with period $T$ :

x(t) = \sum_{k=-\infty}^{\infty} a_k\, e^{jk\omega_0 t} = \sum_{k=-\infty}^{\infty} a_k\, e^{jk(2\pi/T)t}

This is the synthesis equation. $a_0$ is the dc term, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$ th harmonics.

Analysis equation

To find $a_k$ , multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period $T$ :

\int_0^T x(t)e^{-jn\omega_0 t}\,dt = \sum_{k=-\infty}^{\infty} a_k \int_0^T e^{j(k-n)\omega_0 t}\,dt

Evaluate the integral on the right (orthogonality):

\int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T \cos((k-n)\omega_0 t)\,dt + j\int_0^T \sin((k-n)\omega_0 t)\,dt

For $k \ne n$ , both integrals cover a whole number $(k-n)$ of periods, so they are 0.
For $k = n$ , the integrand is 1 and the integral is $T$ .

\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}

So only the $k = n$ term survives:

\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T \quad\Rightarrow\quad a_n = \frac{1}{T}\int_0^T x(t)e^{-jn\omega_0 t}dt

Since any interval of length $T$ gives the same result, the analysis equation is

a_k = \frac{1}{T}\int_{T} x(t)\,e^{-jk\omega_0 t}\,dt

The Fourier series pair

x(t) \overset{FS}{\longleftrightarrow} a_k

Equation	Formula
Synthesis	$x(t) = \sum_k a_k e^{jk\omega_0 t}$
Analysis	$a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$
DC value	$a_0 = \frac{1}{T}\int_T x(t)\,dt$ (average value)

Notes:

The $a_k$ are the Fourier series coefficients or spectral coefficients; they are complex in general, $a_k = |a_k|e^{j\angle a_k}$ , giving the magnitude and phase spectra.
For real $x(t)$ , $a_{-k} = a_k^*$ , so $|a_k|$ is even and $\angle a_k$ is odd in $k$ .

Example: $x(t) = \cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$ gives $a_1 = a_{-1} = 1/2$ and all other $a_k = 0$ .

Asked 8 times
2082 Kartik · 3 marks
2079 Jestha · 6 marks
2078 Baisakh · 6 marks
2076 Bhadra · 5 marks
2073 Bhadra · 5 marks
2072 Magh · 2+4 marks
2072 Asoj · 5 marks
2083 Bhadra (new course) · 4 marks

State and prove Parseval's relation for Discrete Time Fourier Series (discrete time periodic signals).

Answer

Parseval's relation for the DTFS: the average power of a discrete-time periodic signal over one period equals the sum of the squared magnitudes of its Fourier series coefficients:

\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2

Here $x[n]$ has period $N$ , $\omega_0 = 2\pi/N$ , and

x[n] = \sum_{k=\langle N\rangle} a_k e^{jk\omega_0 n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk\omega_0 n}

Proof

Start from the left side and write $|x[n]|^2 = x[n]\,x^*[n]$ :

\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,x^*[n]

Replace $x^*[n]$ by the conjugate of the synthesis equation, $x^*[n] = \sum_{k=\langle N\rangle}a_k^* e^{-jk\omega_0 n}$ :

\begin{aligned} \frac{1}{N}\sum_{n}|x[n]|^2 &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\sum_{k=\langle N\rangle}a_k^* e^{-jk\omega_0 n} \\ &= \sum_{k=\langle N\rangle}a_k^*\left[\frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk\omega_0 n}\right] \\ &= \sum_{k=\langle N\rangle}a_k^*\, a_k \\ &= \sum_{k=\langle N\rangle}|a_k|^2 \end{aligned}

The bracket is exactly the analysis equation for $a_k$ , which completes the proof.

Meaning

$|a_k|^2$ is the average power in the $k$ th harmonic $a_k e^{jk\omega_0 n}$ , because $|e^{jk\omega_0 n}| = 1$ .
So total average power = sum of powers of the $N$ harmonic components; power can be computed in either the time or the frequency domain.
Only $N$ distinct coefficients exist ( $a_{k+N} = a_k$ ), so the sum is finite.

Example: $x[n] = \cos(\pi n/2)$ , $N = 4$ : $a_1 = a_{-1} = a_3 = 1/2$ . Time domain: samples $1, 0, -1, 0$ give $\frac{1}{4}(1 + 0 + 1 + 0) = 0.5$ . Frequency domain: $|a_1|^2 + |a_{-1}|^2 = 0.25 + 0.25 = 0.5$ . Both agree.

Asked 6 times
2077 Chaitra · 4 marks
2076 Baisakh · 6 marks
2075 Bhadra · 6+4 marks
2072 Magh · 5 marks
2071 Magh · 7 marks
2083 Baisakh (new course) · 4 marks

Derive the exponential form of the Fourier series (synthesis and analysis equations) for a discrete time periodic signal x[n] with period N.

Answer

A discrete-time signal $x[n]$ is periodic with period $N$ if $x[n+N] = x[n]$ ; its fundamental frequency is $\omega_0 = 2\pi/N$ . The DTFS writes it as a sum of harmonically related complex exponentials $\phi_k[n] = e^{jk\omega_0 n}$ .

Only $N$ distinct exponentials

\phi_{k+N}[n] = e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n}\, e^{j2\pi n} = \phi_k[n]

since $e^{j2\pi n} = 1$ for integer $n$ . So only $N$ different exponentials exist, e.g. $k = 0, 1, \dots, N-1$ . The sum therefore runs over any $N$ consecutive values of $k$ , written $k = \langle N\rangle$ .

Synthesis equation

x[n] = \sum_{k=\langle N\rangle} a_k\, e^{jk\omega_0 n} = \sum_{k=\langle N\rangle} a_k\, e^{jk(2\pi/N)n}

Orthogonality result

For integer $r$ , using the finite geometric series:

\sum_{n=0}^{N-1} e^{jr(2\pi/N)n} = \begin{cases} N, & r = 0, \pm N, \pm 2N, \dots \\ \dfrac{1 - e^{j2\pi r}}{1 - e^{j2\pi r/N}} = 0, & \text{otherwise} \end{cases}

Analysis equation

Multiply the synthesis equation by $e^{-jr(2\pi/N)n}$ and sum over one period:

\begin{aligned} \sum_{n=\langle N\rangle}x[n]e^{-jr(2\pi/N)n} &= \sum_{n=\langle N\rangle}\sum_{k=\langle N\rangle}a_k e^{j(k-r)(2\pi/N)n} \\ &= \sum_{k=\langle N\rangle}a_k \sum_{n=\langle N\rangle}e^{j(k-r)(2\pi/N)n} \end{aligned}

The inner sum is $N$ when $k = r$ (within one period of $k$ ) and 0 otherwise, so

\sum_{n=\langle N\rangle}x[n]e^{-jr(2\pi/N)n} = N a_r

a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}

The DTFS pair

Equation	Formula
Synthesis	$x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$
Analysis	$a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk(2\pi/N)n}$

Key points:

Both sums are finite ( $N$ terms), so there is no convergence problem, unlike the CT Fourier series.
The coefficients are periodic: $a_{k+N} = a_k$ .
$a_0 = \frac{1}{N}\sum x[n]$ is the average (dc) value.

Example: $x[n] = \sin(2\pi n/5)$ , $N = 5$ : $x[n] = \frac{1}{2j}e^{j(2\pi/5)n} - \frac{1}{2j}e^{-j(2\pi/5)n}$ , so $a_1 = \frac{1}{2j}$ , $a_{-1} = -\frac{1}{2j}$ , and these repeat every 5 values of $k$ .

Asked 4 times
2082 Chaitra · 4 marks
2078 Chaitra · 2+3 marks
2076 Baisakh · 2+4 marks
2071 Bhadra · 6 marks

What is Parseval's relation? State and prove Parseval's relation for continuous-time periodic signals.

Answer

Parseval's relation states that the average power of a periodic signal, computed in the time domain over one period, equals the sum of the powers of all its harmonic components (the squared magnitudes of its Fourier series coefficients). Power is the same whether measured in time or frequency.

Statement (continuous-time periodic signals)

If $x(t)$ has period $T$ and $x(t) \overset{FS}{\longleftrightarrow} a_k$ , then

\frac{1}{T}\int_{T}|x(t)|^2\,dt = \sum_{k=-\infty}^{\infty}|a_k|^2

Proof

With $\omega_0 = 2\pi/T$ , $x(t) = \sum_k a_k e^{jk\omega_0 t}$ and $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ .

\begin{aligned} \frac{1}{T}\int_T |x(t)|^2dt &= \frac{1}{T}\int_T x(t)\,x^*(t)\,dt \\ &= \frac{1}{T}\int_T x(t)\left[\sum_{k=-\infty}^{\infty}a_k^* e^{-jk\omega_0 t}\right]dt \\ &= \sum_{k=-\infty}^{\infty}a_k^*\left[\frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt\right] \\ &= \sum_{k=-\infty}^{\infty}a_k^*\,a_k = \sum_{k=-\infty}^{\infty}|a_k|^2 \end{aligned}

(the order of sum and integral is swapped, and the bracket is the analysis equation).

Interpretation

$|a_k|^2$ is the average power of the $k$ th harmonic, since $\frac{1}{T}\int_T |a_k e^{jk\omega_0 t}|^2dt = |a_k|^2$ .
A plot of $|a_k|^2$ against $k\omega_0$ is the power spectrum of the signal.

Example: $x(t) = A\cos\omega_0 t$ has $a_{\pm1} = A/2$ . Frequency domain: $(A/2)^2 + (A/2)^2 = A^2/2$ ; time domain: $\frac{1}{T}\int_T A^2\cos^2\omega_0 t\,dt = A^2/2$ . They match.

Asked 3 times
2081 Chaitra · 5 marks
2079 Asoj · 5 marks
2078 Chaitra · 5 marks

Derive the trigonometric Fourier series expansion of a continuous-time periodic signal x(t).

Answer

A periodic signal $x(t)$ with period $T$ ( $\omega_0 = 2\pi/T$ ) that meets the Dirichlet conditions can be written as a dc term plus sums of cosines and sines at multiples of $\omega_0$ . This is the trigonometric Fourier series:

x(t) = a_0 + \sum_{n=1}^{\infty}\left[a_n\cos n\omega_0 t + b_n \sin n\omega_0 t\right]

Orthogonality relations (over one period $T$ )

For integers $m, n \ge 1$ :

\begin{aligned} &\int_T \cos m\omega_0 t\,dt = \int_T \sin m\omega_0 t\,dt = 0 \\ &\int_T \cos m\omega_0 t\,\cos n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\ &\int_T \sin m\omega_0 t\,\sin n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\ &\int_T \sin m\omega_0 t\,\cos n\omega_0 t\,dt = 0 \quad \text{for all } m, n \end{aligned}

Finding $a_0$

Integrate both sides over one period. Every sine and cosine term integrates to zero:

\int_T x(t)\,dt = a_0 T \quad\Rightarrow\quad a_0 = \frac{1}{T}\int_T x(t)\,dt

Finding $a_n$

Multiply both sides by $\cos m\omega_0 t$ and integrate over $T$ . Only the term with $n = m$ in the cosine sum survives:

\int_T x(t)\cos m\omega_0 t\,dt = a_m\frac{T}{2} \quad\Rightarrow\quad a_n = \frac{2}{T}\int_T x(t)\cos n\omega_0 t\,dt

Finding $b_n$

Multiply by $\sin m\omega_0 t$ and integrate; only the $n = m$ sine term survives:

\int_T x(t)\sin m\omega_0 t\,dt = b_m\frac{T}{2} \quad\Rightarrow\quad b_n = \frac{2}{T}\int_T x(t)\sin n\omega_0 t\,dt

Summary and notes

Coefficient	Formula
$a_0$ (dc)	$\frac{1}{T}\int_T x(t)\,dt$
$a_n$	$\frac{2}{T}\int_T x(t)\cos n\omega_0 t\,dt$
$b_n$	$\frac{2}{T}\int_T x(t)\sin n\omega_0 t\,dt$

Even $x(t)$ : $b_n = 0$ (cosine terms only). Odd $x(t)$ : $a_0 = a_n = 0$ (sine terms only).
Compact (polar) form: $x(t) = C_0 + \sum_{n=1}^{\infty}C_n\cos(n\omega_0 t + \theta_n)$ with $C_0 = a_0$ , $C_n = \sqrt{a_n^2 + b_n^2}$ , $\theta_n = -\tan^{-1}(b_n/a_n)$ .
Link to exponential form: $c_0 = a_0$ , $c_n = \frac{a_n - jb_n}{2}$ , $c_{-n} = \frac{a_n + jb_n}{2}$ .

Asked 3 times
2082 Kartik · 5 marks
2083 Bhadra (new course) · 5 marks
2083 Baisakh (new course) · 4 marks

Find and plot the Fourier series coefficients (magnitude and associated phase) for the following signal with period N. x[n] = 1 + sin((2π/N)n) + 3cos((2π/N)n) + cos((4π/N)n + π/2)

Answer

Fundamental frequency $\omega_0 = 2\pi/N$ . Expand each term with Euler's formula and read off the coefficients of $e^{jk\omega_0 n}$ :

\begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\omega_0 n} - e^{-j\omega_0 n}\right] + \frac{3}{2}\left[e^{j\omega_0 n} + e^{-j\omega_0 n}\right] \\ &\quad + \frac{1}{2}\left[e^{j(2\omega_0 n + \pi/2)} + e^{-j(2\omega_0 n + \pi/2)}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j\omega_0 n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j\omega_0 n} \\ &\quad + \frac{1}{2}e^{j\pi/2}e^{j2\omega_0 n} + \frac{1}{2}e^{-j\pi/2}e^{-j2\omega_0 n} \end{aligned}

Fourier series coefficients (one period, $k = -2$ to $2$ )

Using $\frac{1}{2j} = -\frac{j}{2}$ :

$k$	$a_k$	$\lvert a_k\rvert$	$\angle a_k$
0	$1$	1	$0$
1	$\frac{3}{2} - \frac{j}{2}$	$\frac{\sqrt{10}}{2} = 1.581$	$-18.43^\circ$
$-1$	$\frac{3}{2} + \frac{j}{2}$	1.581	$+18.43^\circ$
2	$\frac{j}{2}$	0.5	$+90^\circ$
$-2$	$-\frac{j}{2}$	0.5	$-90^\circ$
others in the period	0	0	-

Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$ , $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$ ( $-0.322$ rad).

The coefficients repeat with period $N$ : $a_{k+N} = a_k$ (e.g. $a_{N-1} = a_{-1}$ , $a_{N-2} = a_{-2}$ ). Here $N \ge 5$ is assumed so that the harmonics do not overlap.

Plots (one period, $k = -2 \dots 2$ , repeating every $N$ )

 |a_k|  (repeats every N)
 1.581     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
    90              |
 18.43     |        |
       -+--+--o--+--+--> k
-18.43  |        |
   -90  |
       -2 -1  0  1  2

Magnitude is even in $k$ and phase is odd in $k$ , as expected for a real signal ( $a_{-k} = a_k^*$ ).

Asked 2 times
2081 Chaitra · 5 marks
2076 Baisakh · 4 marks

State and prove the time-scaling property for continuous-time (exponential) Fourier series pair.

Answer

Statement

If $x(t)$ is periodic with period $T$ (fundamental frequency $\omega_0 = 2\pi/T$ ) and

x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t},

then for $\alpha > 0$ , the time-scaled signal $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$ , and its Fourier series coefficients are the same $a_k$ :

x(\alpha t) = \sum_{k=-\infty}^{\infty}a_k\, e^{jk(\alpha\omega_0)t}

Time scaling does not change the coefficients; it only changes the spacing of the spectral lines from $\omega_0$ to $\alpha\omega_0$ .

Proof

Synthesis route: replace $t$ by $\alpha t$ in the synthesis equation:

x(\alpha t) = \sum_{k}a_k e^{jk\omega_0(\alpha t)} = \sum_k a_k e^{jk(\alpha\omega_0)t}

This is a Fourier series with fundamental frequency $\alpha\omega_0$ and coefficients $a_k$ .

Analysis route (check): let $y(t) = x(\alpha t)$ , period $T' = T/\alpha$ , $\omega_0' = \alpha\omega_0$ :

\begin{aligned} b_k &= \frac{1}{T'}\int_0^{T'} x(\alpha t)e^{-jk\alpha\omega_0 t}dt \qquad (\text{put } \tau = \alpha t,\ dt = d\tau/\alpha) \\ &= \frac{\alpha}{T}\int_0^{T}x(\tau)e^{-jk\omega_0\tau}\frac{d\tau}{\alpha} \\ &= \frac{1}{T}\int_0^T x(\tau)e^{-jk\omega_0\tau}d\tau = a_k \end{aligned}

Hence $b_k = a_k$ .

Interpretation

Compressing in time ( $\alpha > 1$ ) spreads the spectral lines further apart in frequency; expanding ( $\alpha < 1$ ) brings them closer.
The amplitude and phase of each harmonic are unchanged, only its frequency $k\omega_0$ becomes $k\alpha\omega_0$ .

Example: $x(t) = \cos(2t)$ has $a_{\pm1} = 1/2$ at $\omega_0 = 2$ . Then $x(3t) = \cos(6t)$ still has $a_{\pm1} = 1/2$ , now at $\omega_0' = 6$ rad/s.

Asked 2 times
2081 Asoj · 6 marks
2082 Bhadra (new course) · 4 marks

Consider a signal x[n] = sin(ω₀n) with period N = 6. Find the Fourier series coefficient of the given signal x[n]. Plot spectra. Also find the average power.

Answer

Since $x[n] = \sin(\omega_0 n)$ has period $N = 6$ , the fundamental frequency is

\omega_0 = \frac{2\pi}{N} = \frac{2\pi}{6} = \frac{\pi}{3}

Fourier series coefficients

Using Euler's formula:

x[n] = \sin\frac{\pi n}{3} = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n}

Comparing with $x[n] = \sum_{k=\langle 6\rangle}a_k e^{jk(2\pi/6)n}$ :

a_1 = \frac{1}{2j} = -\frac{j}{2} = \frac{1}{2}e^{-j\pi/2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2} = \frac{1}{2}e^{j\pi/2}

All other $a_k$ in one period are zero. The coefficients repeat every 6: $a_{-1} = a_5 = a_{11}$ , $a_1 = a_7 = a_{-5}$ , etc.

$k$	0	1	2	3	4	5
$a_k$	0	$-j/2$	0	0	0	$j/2$
$\lvert a_k\rvert$	0	0.5	0	0	0	0.5
$\angle a_k$	-	$-90^\circ$	-	-	-	$+90^\circ$

Spectra

 |a_k|  (repeats every 6)
   0.5  |           |     |           |
       -+--o--o--o--+--o--+--o--o--o--+--> k
       -5 -4 -3 -2 -1  0  1  2  3  4  5

 angle a_k (deg)
    90              |                 |
       -+--o--o--o--+--o--+--o--o--o--+--> k
   -90  |                 |
       -5 -4 -3 -2 -1  0  1  2  3  4  5

Lines of height $1/2$ at $k = \pm1, \pm5, \pm7, \dots$ ; phase $-90^\circ$ at $k = 1, 7, -5, \dots$ and $+90^\circ$ at $k = -1, 5, \dots$

Average power

By Parseval's relation:

P = \sum_{k=\langle 6\rangle}|a_k|^2 = |a_1|^2 + |a_5|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Check in time: samples over one period are $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$ , so $P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = 0.5$ .

Answer: $a_1 = -j/2$ , $a_{-1} = a_5 = j/2$ , others 0; average power $P = 0.5$ W.

Asked 2 times
2079 Asoj · 4 marks
2079 Chaitra · 4 marks

State and prove the time-scaling property for discrete-time Fourier series pair.

Answer

In discrete time, $x[an]$ is not defined for non-integer $an$ , so time scaling is defined through the time-expanded signal (inserting zeros).

Statement

Let $x[n]$ be periodic with period $N$ and DTFS coefficients $a_k$ . For a positive integer $m$ define

x_{(m)}[n] = \begin{cases} x[n/m], & n \text{ a multiple of } m \\ 0, & \text{otherwise} \end{cases}

( $m - 1$ zeros are inserted between samples). Then $x_{(m)}[n]$ is periodic with period $mN$ and

x_{(m)}[n] \overset{FS}{\longleftrightarrow} \frac{1}{m}a_k

(with $a_k$ viewed as periodic with period $N$ , so the $mN$ coefficients are $a_k/m$ for $k = 0, 1, \dots, mN-1$ ).

Proof

The period of $x_{(m)}[n]$ is $mN$ , so its coefficients are

b_k = \frac{1}{mN}\sum_{n=0}^{mN-1}x_{(m)}[n]e^{-jk\frac{2\pi}{mN}n}

Only $n = mr$ ( $r = 0, 1, \dots, N-1$ ) give non-zero terms, where $x_{(m)}[mr] = x[r]$ :

\begin{aligned} b_k &= \frac{1}{mN}\sum_{r=0}^{N-1}x[r]\,e^{-jk\frac{2\pi}{mN}mr} \\ &= \frac{1}{m}\left[\frac{1}{N}\sum_{r=0}^{N-1}x[r]e^{-jk\frac{2\pi}{N}r}\right] \\ &= \frac{1}{m}a_k \end{aligned}

Interpretation

Expanding in time by $m$ multiplies the period by $m$ , so the fundamental frequency becomes $\omega_0/m = 2\pi/(mN)$ and the spectrum is packed $m$ times more densely.
The coefficient values are divided by $m$ , because the same energy per period is now averaged over $m$ times as many samples.
The coefficient pattern of $x[n]$ is repeated $m$ times within one period of $b_k$ .

Example: $x[n] = \{1, 1\}$ with $N = 2$ has $a_0 = 1$ , $a_1 = 0$ . With $m = 2$ , $x_{(2)}[n] = \{1, 0, 1, 0\}$ ( $N = 4$ ) gives $b_k = \frac{1}{4}(1 + e^{-j\pi k}) = \{0.5, 0, 0.5, 0\}$ , which is $\frac{1}{2}\{a_0, a_1, a_0, a_1\}$ .

Asked 2 times
2079 Asoj · 6 marks
2079 Chaitra · 6 marks

Find the frequency domain representation (both magnitude and phase plot) of a periodic signal defined as x[n] = {1, −0.5, 0, 0, 0.5} (x[0] = 1), with period N = 5.

Answer

The signal has period $N = 5$ , so $\omega_0 = 2\pi/5$ and

a_k = \frac{1}{5}\sum_{n=0}^{4}x[n]\,e^{-jk(2\pi/5)n}

One period: $x[0] = 1$ , $x[1] = -0.5$ , $x[2] = 0$ , $x[3] = 0$ , $x[4] = 0.5$ . Because of periodicity $x[4] = x[-1]$ , so it is easier to sum over $n = -1, 0, 1$ (note $x[-1] = 0.5$ ):

\begin{aligned} a_k &= \frac{1}{5}\left[x[-1]e^{jk2\pi/5} + x[0] + x[1]e^{-jk2\pi/5}\right] \\ &= \frac{1}{5}\left[1 + 0.5e^{jk2\pi/5} - 0.5e^{-jk2\pi/5}\right] \\ &= \frac{1}{5}\left[1 + 0.5\,(2j)\sin\frac{2\pi k}{5}\right] \\ &= \frac{1}{5}\left[1 + j\sin\frac{2\pi k}{5}\right] \end{aligned}

Magnitude and phase:

|a_k| = \frac{1}{5}\sqrt{1 + \sin^2\frac{2\pi k}{5}}, \qquad \angle a_k = \tan^{-1}\left(\sin\frac{2\pi k}{5}\right)

Values over one period

$k$	$\sin(2\pi k/5)$	$a_k$	$\lvert a_k\rvert$	$\angle a_k$
0	0	$0.2$	0.200	$0^\circ$
1	0.951	$0.2 + j0.190$	0.276	$43.56^\circ$
2	0.588	$0.2 + j0.118$	0.232	$30.45^\circ$
3	$-0.588$	$0.2 - j0.118$	0.232	$-30.45^\circ$
4	$-0.951$	$0.2 - j0.190$	0.276	$-43.56^\circ$

(Since $a_{k+5} = a_k$ : $a_{-1} = a_4$ , $a_{-2} = a_3$ .)

Magnitude and phase plots (repeat every 5)

 |a_k|  (repeats every 5)
 0.276     |     |
 0.232  |  |     |  |
   0.2  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
 43.56           |
 30.45           |  |
       -+--+--o--+--+--> k
-30.45  |  |
-43.56     |
       -2 -1  0  1  2

The magnitude spectrum is even and the phase spectrum is odd in $k$ , as expected for a real $x[n]$ .

Check (synthesis at $n = 0$ ): $\sum_{k=0}^{4}a_k = 5\times 0.2 + j(0) = 1 = x[0]$ . Correct.

Asked 2 times
2080 Asoj · 6 marks
2070 Magh · 6 marks

Find the Fourier series coefficients of the signal x[n] = 1 + sin[(2π/N)n] + 3cos[(2π/N)n] + cos[(4π/N)n]

Answer

Fundamental frequency $\omega_0 = 2\pi/N$ . Rewrite each term with Euler's formula:

\begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\frac{2\pi}{N}n} - e^{-j\frac{2\pi}{N}n}\right] + \frac{3}{2}\left[e^{j\frac{2\pi}{N}n} + e^{-j\frac{2\pi}{N}n}\right] + \frac{1}{2}\left[e^{j\frac{4\pi}{N}n} + e^{-j\frac{4\pi}{N}n}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j(2\pi/N)n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j(2\pi/N)n} + \frac{1}{2}e^{j2(2\pi/N)n} + \frac{1}{2}e^{-j2(2\pi/N)n} \end{aligned}

Comparing with $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$ (and using $\frac{1}{2j} = -\frac{j}{2}$ ):

$k$	$a_k$	$\lvert a_k\rvert$	$\angle a_k$
0	$1$	1	$0$
1	$\frac{3}{2} - \frac{j}{2}$	$\frac{\sqrt{10}}{2} = 1.581$	$-18.43^\circ$
$-1$	$\frac{3}{2} + \frac{j}{2}$	1.581	$+18.43^\circ$
2	$\frac{1}{2}$	0.5	$0$
$-2$	$\frac{1}{2}$	0.5	$0$
other $k$ in one period	0	0	-

Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$ ; $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$ .

The coefficients are periodic with period $N$ : $a_{k+N} = a_k$ , so for example $a_{N-1} = a_{-1}$ and $a_{N-2} = a_{-2}$ (assuming $N \ge 5$ so that the harmonics do not overlap).

 |a_k|  (repeats every N)
 1.581     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

Since $x[n]$ is real, $a_{-k} = a_k^*$ : the magnitude is even and the phase is odd.

Asked 2 times
2078 Chaitra · 4 marks
2075 Baisakh · 4 marks

State and prove the time-shifting property for discrete-time Fourier series pair.

Answer

Statement

If $x[n]$ is periodic with period $N$ and $x[n] \overset{FS}{\longleftrightarrow} a_k$ , then

x[n - n_0] \overset{FS}{\longleftrightarrow} a_k\, e^{-jk(2\pi/N)n_0}

A shift in time multiplies each coefficient by a linear phase term; the magnitudes $|a_k|$ do not change.

Proof

$x[n - n_0]$ is also periodic with period $N$ . Its coefficients are

b_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n - n_0]\,e^{-jk(2\pi/N)n}

Put $m = n - n_0$ , so $n = m + n_0$ . As $n$ runs over one period, $m$ also runs over one period:

\begin{aligned} b_k &= \frac{1}{N}\sum_{m=\langle N\rangle}x[m]\,e^{-jk(2\pi/N)(m + n_0)} \\ &= e^{-jk(2\pi/N)n_0}\left[\frac{1}{N}\sum_{m=\langle N\rangle}x[m]e^{-jk(2\pi/N)m}\right] \\ &= e^{-jk(2\pi/N)n_0}\,a_k \end{aligned}

Interpretation

$|b_k| = |a_k|$ : the magnitude spectrum is unchanged by a shift.
$\angle b_k = \angle a_k - k\frac{2\pi}{N}n_0$ : the phase changes linearly with $k$ .

Example: $x[n] = \cos(2\pi n/N)$ has $a_{\pm1} = 1/2$ . Then $x[n-1]$ has $b_{1} = \frac{1}{2}e^{-j2\pi/N}$ and $b_{-1} = \frac{1}{2}e^{j2\pi/N}$ , still of magnitude $1/2$ .

Asked 2 times
2074 Bhadra · 4 marks
2083 Bhadra (new course) · 4 marks

State and prove time shifting and time scaling properties of continuous time Fourier series.

Answer

Let $x(t)$ be periodic with period $T$ , $\omega_0 = 2\pi/T$ , and $x(t) \overset{FS}{\longleftrightarrow} a_k$ , where $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ .

Time shifting property

Statement:

x(t - t_0) \overset{FS}{\longleftrightarrow} a_k\, e^{-jk\omega_0 t_0}

Proof: $y(t) = x(t - t_0)$ has the same period $T$ . Its coefficients are

\begin{aligned} b_k &= \frac{1}{T}\int_T x(t - t_0)e^{-jk\omega_0 t}dt \qquad (\tau = t - t_0) \\ &= \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0(\tau + t_0)}d\tau \\ &= e^{-jk\omega_0 t_0}\,\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = e^{-jk\omega_0 t_0}\,a_k \end{aligned}

So $|b_k| = |a_k|$ (magnitude unchanged) and the phase changes by $-k\omega_0 t_0$ .

Time scaling property

Statement: for $\alpha > 0$ , $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$ , and

x(\alpha t) = \sum_{k=-\infty}^{\infty}a_k e^{jk(\alpha\omega_0)t}

The coefficients stay $a_k$ ; only the fundamental frequency changes.

Proof: let $y(t) = x(\alpha t)$ , $T' = T/\alpha$ :

\begin{aligned} b_k &= \frac{1}{T'}\int_0^{T'}x(\alpha t)e^{-jk(\alpha\omega_0)t}dt \qquad (\tau = \alpha t,\ dt = d\tau/\alpha) \\ &= \frac{\alpha}{T}\int_0^{T}x(\tau)e^{-jk\omega_0\tau}\frac{d\tau}{\alpha} = a_k \end{aligned}

Summary

Property	Signal	Coefficients	Fundamental
Time shift	$x(t - t_0)$	$a_k e^{-jk\omega_0 t_0}$	$\omega_0$
Time scale	$x(\alpha t)$	$a_k$	$\alpha\omega_0$

Example: $x(t) = \cos\omega_0 t$ ( $a_{\pm1} = 1/2$ ); $x(t - t_0)$ has $a_{\pm1} = \frac{1}{2}e^{\mp j\omega_0 t_0}$ ; $x(2t) = \cos 2\omega_0 t$ still has $a_{\pm1} = 1/2$ but at fundamental $2\omega_0$ .

2082 Chaitra · 6 marks

Describe the time shifting and frequency shifting properties of continuous time Fourier series. Find Fourier series coefficient of signal x(t) = A cos(ω₀t).

Answer

Let $x(t)$ have period $T$ , $\omega_0 = 2\pi/T$ , and $x(t) \overset{FS}{\longleftrightarrow} a_k$ .

Time shifting property

x(t - t_0) \overset{FS}{\longleftrightarrow} a_k e^{-jk\omega_0 t_0}

Proof sketch: with $\tau = t - t_0$ ,

b_k = \frac{1}{T}\int_T x(t - t_0)e^{-jk\omega_0 t}dt = e^{-jk\omega_0 t_0}\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = e^{-jk\omega_0 t_0}a_k

A delay leaves $|a_k|$ unchanged and adds a phase $-k\omega_0 t_0$ that grows linearly with $k$ .

Frequency shifting property

e^{jM\omega_0 t}\,x(t) \overset{FS}{\longleftrightarrow} a_{k-M}

Proof:

b_k = \frac{1}{T}\int_T e^{jM\omega_0 t}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(t)e^{-j(k-M)\omega_0 t}dt = a_{k-M}

Multiplying by a complex exponential at the $M$ th harmonic shifts the whole set of coefficients by $M$ places (this is the basis of modulation).

Fourier series coefficients of $x(t) = A\cos(\omega_0 t)$

Using Euler's formula:

x(t) = A\cos\omega_0 t = \frac{A}{2}e^{j\omega_0 t} + \frac{A}{2}e^{-j\omega_0 t}

Comparing with $x(t) = \sum_k a_k e^{jk\omega_0 t}$ :

a_1 = a_{-1} = \frac{A}{2}, \qquad a_k = 0 \text{ for } k \ne \pm1

Check with the analysis equation for $k = 1$ :

a_1 = \frac{1}{T}\int_0^T A\cos\omega_0 t\,e^{-j\omega_0 t}dt = \frac{A}{2T}\int_0^T\left(1 + e^{-j2\omega_0 t}\right)dt = \frac{A}{2}

 a_k
 A/2      |       |
      ----+---+---+----> k
         -1   0   1

Answer: $a_{\pm1} = A/2$ (real, zero phase), all other coefficients are zero.

2081 Chaitra · 3 marks

State and prove frequency shifting property of discrete time Fourier series.

Answer

Statement

If $x[n]$ is periodic with period $N$ and $x[n] \overset{FS}{\longleftrightarrow} a_k$ , then for an integer $M$ :

e^{jM(2\pi/N)n}\,x[n] \overset{FS}{\longleftrightarrow} a_{k-M}

Multiplying by a complex exponential at the $M$ th harmonic shifts the coefficient sequence by $M$ .

Proof

$y[n] = e^{jM(2\pi/N)n}x[n]$ is also periodic with period $N$ (both factors are). Its coefficients are

\begin{aligned} b_k &= \frac{1}{N}\sum_{n=\langle N\rangle}e^{jM(2\pi/N)n}x[n]\,e^{-jk(2\pi/N)n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k-M)(2\pi/N)n} \\ &= a_{k-M} \end{aligned}

Example: for $N = 4$ , multiplying by $e^{j\pi n/2}$ ( $M = 1$ ) moves $a_0$ to position $k = 1$ , $a_1$ to $k = 2$ , and so on (cyclically, since $a_k$ has period 4).

2081 Asoj · 6 marks

What is Fourier series representation of a signal? Derive an expression for analysis and synthesis equation of the Fourier Series for a CT periodic signal x(t).

Answer

The Fourier series representation of a periodic signal expresses it as a sum (linear combination) of harmonically related sinusoids or complex exponentials, whose frequencies are integer multiples of the fundamental frequency $\omega_0 = 2\pi/T$ . It shows which frequencies the signal contains and how strongly.

Synthesis equation

Each $e^{jk\omega_0 t}$ , $k = 0, \pm1, \pm2, \dots$ , is periodic with period $T$ , so their combination is also periodic with $T$ :

x(t) = \sum_{k=-\infty}^{\infty}a_k\,e^{jk\omega_0 t}

This builds (synthesises) $x(t)$ from its harmonics; $a_k$ are the Fourier series coefficients.

Analysis equation

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

For $k \ne n$ , $e^{j(k-n)\omega_0 t}$ completes $(k - n)$ whole cycles in $T$ , so

\int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T\cos((k-n)\omega_0 t)dt + j\int_0^T\sin((k-n)\omega_0 t)dt = 0

For $k = n$ the integrand is 1 and the integral is $T$ . So only $k = n$ survives:

\int_0^T x(t)e^{-jn\omega_0 t}dt = T a_n

a_k = \frac{1}{T}\int_T x(t)\,e^{-jk\omega_0 t}\,dt

This extracts (analyses) the amount of the $k$ th harmonic in $x(t)$ ; the integral may be taken over any interval of length $T$ .

Summary

Name	Equation
Synthesis	$x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}$
Analysis	$a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$
DC term	$a_0 = \frac{1}{T}\int_T x(t)dt$

Example: $x(t) = \sin\omega_0 t = \frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t}$ , so $a_1 = \frac{1}{2j}$ , $a_{-1} = -\frac{1}{2j}$ , all others 0.

2080 Asoj · 2+6 marks

What are the significance of Fourier series coefficients of a signal? State and prove time scaling and conjugation properties of continuous time Fourier series representation.

Answer

Significance of Fourier series coefficients

The coefficients $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ describe the signal in the frequency domain:

$|a_k|$ gives the amplitude and $\angle a_k$ the phase of the $k$ th harmonic (frequency $k\omega_0$ ); plotting them gives the magnitude and phase spectra.
$a_0$ is the dc (average) value of the signal.
$|a_k|^2$ is the power in the $k$ th harmonic, and by Parseval $\sum|a_k|^2$ is the total average power.
They show the bandwidth of the signal: how fast $|a_k|$ falls tells how many harmonics are needed (smooth signals need fewer).
An LTI system acts on each harmonic separately: the output coefficients are $a_k H(jk\omega_0)$ , which makes system analysis simple.

Time scaling property

Statement: if $x(t)$ has period $T$ , $\omega_0 = 2\pi/T$ , and $x(t) \overset{FS}{\longleftrightarrow} a_k$ , then for $\alpha > 0$ , $x(\alpha t)$ has period $T/\alpha$ and

x(\alpha t) = \sum_{k=-\infty}^{\infty}a_k e^{jk(\alpha\omega_0)t}

i.e. the coefficients are unchanged but the fundamental frequency becomes $\alpha\omega_0$ .

Proof: let $y(t) = x(\alpha t)$ , period $T' = T/\alpha$ ; put $\tau = \alpha t$ :

\begin{aligned} b_k &= \frac{1}{T'}\int_0^{T'}x(\alpha t)e^{-jk\alpha\omega_0 t}dt \\ &= \frac{\alpha}{T}\int_0^{T}x(\tau)e^{-jk\omega_0\tau}\frac{d\tau}{\alpha} = a_k \end{aligned}

Conjugation property

Statement:

x^*(t) \overset{FS}{\longleftrightarrow} a_{-k}^*

Proof: take the conjugate of the synthesis equation:

\begin{aligned} x^*(t) &= \left[\sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}\right]^* = \sum_{k=-\infty}^{\infty}a_k^* e^{-jk\omega_0 t} \\ &= \sum_{m=-\infty}^{\infty}a_{-m}^*\,e^{jm\omega_0 t} \qquad (m = -k) \end{aligned}

So the coefficient of $e^{jm\omega_0 t}$ in $x^*(t)$ is $a_{-m}^*$ .

Consequence (conjugate symmetry): if $x(t)$ is real, $x(t) = x^*(t)$ , so $a_k = a_{-k}^*$ . Hence $|a_k| = |a_{-k}|$ (even magnitude spectrum) and $\angle a_{-k} = -\angle a_k$ (odd phase spectrum). If $x(t)$ is also even, $a_k$ is real and even.

2079 Jestha · 8 marks

Calculate and plot magnitude and phase spectrum of the following periodic signal. [Figure: sawtooth wave of period 1: x(t) rises linearly from 0 at t = 0 to 1 at t = 1 and then drops back to 0, i.e. x(t) = t for 0 ≤ t < 1, repeating every 1 s; shown from t = −3 to t = 4]

Answer

One period: $x(t) = t$ for $0 \le t < 1$ , with period $T = 1$ s, so $\omega_0 = 2\pi/T = 2\pi$ rad/s.

a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \int_0^1 t\,e^{-j2\pi kt}dt

DC term ( $k = 0$ )

a_0 = \int_0^1 t\,dt = \frac{1}{2}

Coefficients for $k \ne 0$

Integrate by parts with $u = t$ , $dv = e^{-j2\pi kt}dt$ , $v = \frac{e^{-j2\pi kt}}{-j2\pi k}$ :

\begin{aligned} a_k &= \left[\frac{t\,e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 - \int_0^1\frac{e^{-j2\pi kt}}{-j2\pi k}dt \\ &= \frac{e^{-j2\pi k}}{-j2\pi k} + \frac{1}{j2\pi k}\left[\frac{e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 \\ &= \frac{1}{-j2\pi k} + \frac{1}{j2\pi k}\cdot\frac{e^{-j2\pi k} - 1}{-j2\pi k} \end{aligned}

Since $e^{-j2\pi k} = 1$ , the second term is zero:

a_k = \frac{-1}{j2\pi k} = \frac{j}{2\pi k}, \qquad k \ne 0

Magnitude and phase

|a_k| = \frac{1}{2\pi|k|}, \qquad \angle a_k = \begin{cases} +90^\circ\ (\pi/2), & k > 0 \\ -90^\circ\ (-\pi/2), & k < 0 \\ 0, & k = 0 \end{cases}

$k$	0	$\pm1$	$\pm2$	$\pm3$	$\pm4$
$\lvert a_k\rvert$	0.5	0.159	0.0796	0.0531	0.0398
$\angle a_k$ ( $k>0$ )	$0^\circ$	$90^\circ$	$90^\circ$	$90^\circ$	$90^\circ$
$\angle a_k$ ( $k<0$ )	-	$-90^\circ$	$-90^\circ$	$-90^\circ$	$-90^\circ$

So the Fourier series is

x(t) = \frac{1}{2} + \sum_{k\ne0}\frac{j}{2\pi k}e^{j2\pi kt} = \frac{1}{2} - \sum_{k=1}^{\infty}\frac{1}{\pi k}\sin(2\pi kt)

Spectra (lines at $\omega = 2\pi k$ )

 |a_k|
   0.5           |
 0.159        |  |  |
0.0796     |  |  |  |  |
0.0531  |  |  |  |  |  |  |
       -+--+--+--+--+--+--+--> k
       -3 -2 -1  0  1  2  3

 angle a_k (deg)
    90              |  |  |
       -+--+--+--o--+--+--+--> k
   -90  |  |  |
       -3 -2 -1  0  1  2  3

The magnitude falls as $1/k$ (because of the jump discontinuity each period), is even in $k$ , and the phase is odd: $+90^\circ$ for positive $k$ and $-90^\circ$ for negative $k$ .

Check: at $t = 0.25$ , the series $\frac{1}{2} - \frac{1}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \dots\right) = \frac{1}{2} - \frac{1}{\pi}\cdot\frac{\pi}{4} = 0.25$ , which equals $x(0.25)$ .

2078 Baisakh · 4 marks

"Even if a signal is periodic, its Fourier series might not exist." Justify the statement using three different examples.

Answer

A periodic signal has a Fourier series that converges to it only if it meets the Dirichlet conditions over one period: (1) absolutely integrable, (2) finite number of maxima and minima, (3) finite number of finite discontinuities. Periodicity alone is not enough. Three periodic signals (period $T = 1$ ) that each break one condition:

Example 1: not absolutely integrable

x(t) = \frac{1}{t}, \quad 0 < t \le 1, \quad \text{repeated with period } 1

\int_0^1\left|\frac{1}{t}\right|dt = \left[\ln t\right]_0^1 = \infty

The coefficients $a_k = \int_0^1 \frac{1}{t}e^{-j2\pi kt}dt$ are not finite, so the series does not exist.

Example 2: infinite number of maxima and minima

x(t) = \sin\left(\frac{2\pi}{t}\right), \quad 0 < t \le 1, \quad \text{period } 1

It is bounded and absolutely integrable ( $\int_0^1|x|\,dt < 1$ ), but as $t \to 0$ it oscillates infinitely fast, giving infinitely many maxima and minima in one period. The series does not converge to $x(t)$ .

Example 3: infinite number of discontinuities

A staircase signal over $0 \le t < 1$ that equals 1 on $[0, 1/2)$ , $1/2$ on $[1/2, 3/4)$ , $1/4$ on $[3/4, 7/8)$ , and so on (each step half the height and half the length of the previous one), repeated every 1 s.

It is bounded and $\int_0^1|x|\,dt < 1$ , but it has an infinite number of discontinuities within one period, so it violates the third condition.

Example	Condition violated
$1/t$ on $(0, 1]$	absolute integrability
$\sin(2\pi/t)$	finite maxima/minima
halving staircase	finite discontinuities

So even though all three are periodic, their Fourier series do not exist (or do not converge to them). Such signals are mathematical curiosities; practical signals usually meet the Dirichlet conditions.

2078 Poush · 5+3 marks

Derive the formula for synthesis equation and analysis equation of Fourier series. Mention the three conditions that guarantee the existence of Fourier series.

Answer

A periodic signal $x(t)$ with period $T$ and fundamental frequency $\omega_0 = 2\pi/T$ can be written as a sum of harmonically related complex exponentials (the exponential Fourier series).

Synthesis equation

The functions $e^{jk\omega_0 t}$ , $k = 0, \pm1, \pm2, \dots$ , all repeat every $T$ , so

x(t) = \sum_{k=-\infty}^{\infty}a_k\,e^{jk\omega_0 t}

is periodic with period $T$ . This is the synthesis equation; $a_k$ are the Fourier series coefficients.

Analysis equation

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

Orthogonality of the exponentials:

\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}

(for $k \ne n$ , the cosine and sine parts complete a whole number of cycles and integrate to zero). Hence

\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T

a_k = \frac{1}{T}\int_T x(t)\,e^{-jk\omega_0 t}\,dt

This is the analysis equation; $a_0 = \frac{1}{T}\int_T x(t)dt$ is the dc value.

Equation	Formula
Synthesis	$x(t) = \sum_k a_k e^{jk\omega_0 t}$
Analysis	$a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$

Conditions for existence (Dirichlet conditions)

Over any one period:

Absolutely integrable: $\int_T |x(t)|\,dt < \infty$ , which guarantees every $|a_k|$ is finite.
Finite number of maxima and minima (bounded variation) in one period.
Finite number of discontinuities in one period, and each discontinuity is finite.

When these hold, the series converges to $x(t)$ at every point where $x(t)$ is continuous, and to the average $\frac{1}{2}[x(t^-) + x(t^+)]$ at a discontinuity. (A weaker condition, finite energy over one period, $\int_T|x(t)|^2dt < \infty$ , guarantees convergence in the mean-square sense.)

2078 Poush · 6 marks

State and explain four properties of discrete time Fourier series.

Answer

Let $x[n]$ and $y[n]$ be periodic with the same period $N$ , $\omega_0 = 2\pi/N$ , with $x[n] \overset{FS}{\longleftrightarrow} a_k$ and $y[n] \overset{FS}{\longleftrightarrow} b_k$ , where $a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk\omega_0 n}$ .

1. Linearity

A\,x[n] + B\,y[n] \overset{FS}{\longleftrightarrow} A a_k + B b_k

The analysis sum is linear, so the coefficients of a weighted sum are the same weighted sum of coefficients. Example: $x[n] = 2\cos(\omega_0 n) + 3$ has $a_0 = 3$ , $a_{\pm1} = 1$ .

2. Time shifting

x[n - n_0] \overset{FS}{\longleftrightarrow} a_k e^{-jk(2\pi/N)n_0}

Proof: with $m = n - n_0$ , $\frac{1}{N}\sum_n x[n-n_0]e^{-jk\omega_0 n} = e^{-jk\omega_0 n_0}\frac{1}{N}\sum_m x[m]e^{-jk\omega_0 m}$ . A shift leaves $|a_k|$ unchanged and adds a linear phase.

3. Frequency shifting

e^{jM(2\pi/N)n}x[n] \overset{FS}{\longleftrightarrow} a_{k-M}

Multiplying by the $M$ th harmonic exponential shifts the coefficients by $M$ (used in modulation).

4. Parseval's relation

\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2

The average power in one period equals the sum of the powers of the $N$ harmonics.

Other properties (for reference)

Property	Signal	Coefficients
Time reversal	$x[-n]$	$a_{-k}$
Conjugation	$x^*[n]$	$a_{-k}^*$
Periodic convolution	$\sum_{r=\langle N\rangle}x[r]y[n-r]$	$N a_k b_k$
Multiplication	$x[n]y[n]$	$\sum_{l=\langle N\rangle}a_l b_{k-l}$
First difference	$x[n] - x[n-1]$	$(1 - e^{-jk\omega_0})a_k$
Real signal	$x[n]$ real	$a_k = a_{-k}^*$

A property special to the DTFS is that the coefficients are themselves periodic: $a_{k+N} = a_k$ .

2080 Chaitra · 3+5 marks

Write down the condition of convergence of continuous time Fourier series. Find the Fourier series coefficient a_k for the continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2.

Answer

Condition of convergence (Dirichlet conditions)

The Fourier series of a periodic $x(t)$ converges if, over one period:

$x(t)$ is absolutely integrable: $\int_T|x(t)|dt < \infty$ .
$x(t)$ has a finite number of maxima and minima.
$x(t)$ has a finite number of finite discontinuities.

Then the series equals $x(t)$ wherever $x(t)$ is continuous and equals $\frac{1}{2}[x(t^-) + x(t^+)]$ at a jump. (Alternatively, finite energy in one period gives convergence in the mean-square sense.)

Fourier series coefficients

Period $T = 2$ s, so $\omega_0 = 2\pi/T = \pi$ rad/s.

x(t) = \begin{cases} 1.5, & 0 \le t < 1 \\ -1.5, & 1 \le t < 2 \end{cases}

DC term:

a_0 = \frac{1}{2}\left[\int_0^1 1.5\,dt + \int_1^2(-1.5)\,dt\right] = \frac{1}{2}(1.5 - 1.5) = 0

For $k \ne 0$ :

\begin{aligned} a_k &= \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right] \\ &= \frac{1.5}{2}\left[\frac{1 - e^{-jk\pi}}{jk\pi} - \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi}\right] \end{aligned}

Using $e^{-j2k\pi} = 1$ and $e^{-jk\pi} = (-1)^k$ :

\begin{aligned} a_k &= \frac{0.75}{jk\pi}\left[1 - (-1)^k - (-1)^k + 1\right] = \frac{0.75 \times 2\left[1 - (-1)^k\right]}{jk\pi} \\ &= \frac{1.5\left[1 - (-1)^k\right]}{jk\pi} \end{aligned}

So

a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even (including } k = 0) \end{cases}

$k$	$\pm1$	$\pm2$	$\pm3$	$\pm5$
$a_k$	$\mp j0.955$	0	$\mp j0.318$	$\mp j0.191$
$\lvert a_k\rvert$	0.955	0	0.318	0.191

Phase: $-90^\circ$ for positive odd $k$ and $+90^\circ$ for negative odd $k$ (the signal is odd, so the $a_k$ are purely imaginary).

Equivalent trigonometric form:

x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]

Answer: $a_0 = 0$ ; $a_k = \dfrac{3}{jk\pi}$ for odd $k$ ; $a_k = 0$ for even $k$ .

2079 Chaitra · 5+3 marks

Derive the expression of Fourier Series Coefficient in exponential form for Continuous time periodic signal. Describe the condition for existence of Fourier series with suitable examples.

Answer

Exponential Fourier series coefficients

Let $x(t)$ be periodic with period $T$ , $\omega_0 = 2\pi/T$ . It is expressed as a sum of harmonically related complex exponentials (synthesis equation):

x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

The right-hand integral:

\int_0^T e^{j(k-n)\omega_0 t}dt = \left[\frac{e^{j(k-n)\omega_0 t}}{j(k-n)\omega_0}\right]_0^T = \frac{e^{j(k-n)2\pi} - 1}{j(k-n)\omega_0} = 0 \quad (k \ne n)

and equals $T$ when $k = n$ . Only the $k = n$ term remains:

\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T

a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt

This is the analysis equation. $a_0 = \frac{1}{T}\int_T x(t)dt$ is the average value, and $|a_k|$ , $\angle a_k$ give the magnitude and phase spectra.

Conditions for existence (Dirichlet conditions)

Over one period:

Absolutely integrable: $\int_T|x(t)|dt < \infty$ $\int_{T} ∣ x (t) ∣ d t < \infty$ .
- Violated by $x(t) = 1/t$ on $0 < t \le 1$ (period 1): $\int_0^1 dt/t = \infty$ .
Finite number of maxima and minima in a period.
- Violated by $x(t) = \sin(2\pi/t)$ on $0 < t \le 1$ : it oscillates infinitely often near $t = 0$ .
Finite number of discontinuities, each of finite size, in a period.
- Violated by a staircase that steps down by half on $[0, \frac12), [\frac12, \frac34), [\frac34, \frac78), \dots$ : infinitely many jumps in one period.

A signal that satisfies them: a square wave has two finite jumps per period, is bounded and has no oscillations, so its Fourier series exists; it converges to the midpoint value at each jump.

2078 Chaitra · 2+3 marks

State and prove convolution property of discrete time periodic signals.

Answer

Statement

Let $x[n]$ and $y[n]$ be periodic with the same period $N$ , with $x[n] \overset{FS}{\longleftrightarrow} a_k$ and $y[n] \overset{FS}{\longleftrightarrow} b_k$ . Their periodic convolution

z[n] = \sum_{r=\langle N\rangle}x[r]\,y[n - r]

is periodic with period $N$ , and

z[n] \overset{FS}{\longleftrightarrow} c_k = N a_k b_k

Convolution in time becomes multiplication of coefficients.

Proof

With $\omega_0 = 2\pi/N$ :

\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}z[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=\langle N\rangle}\sum_{r=\langle N\rangle}x[r]\,y[n-r]\,e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{r=\langle N\rangle}x[r]\,e^{-jk\omega_0 r}\sum_{n=\langle N\rangle}y[n-r]\,e^{-jk\omega_0(n-r)} \end{aligned}

Put $m = n - r$ ; since $y$ is periodic, $m$ still runs over one full period, so the inner sum is $\sum_{m=\langle N\rangle}y[m]e^{-jk\omega_0 m} = N b_k$ . Then

c_k = \frac{1}{N}\sum_{r=\langle N\rangle}x[r]e^{-jk\omega_0 r}\cdot N b_k = N\,a_k\,b_k

Hence $\sum_{r=\langle N\rangle}x[r]y[n-r] \overset{FS}{\longleftrightarrow} N a_k b_k$ . (The dual property: multiplication $x[n]y[n]$ gives the periodic convolution of coefficients $\sum_{l=\langle N\rangle}a_l b_{k-l}$ .)

2077 Chaitra · 2 marks

A continuous time Fourier Series is evaluated over infinite duration while a discrete time Fourier series is evaluated over one time period only. Why?

Answer

The difference comes from how many distinct harmonics each signal can have:

Continuous time: the exponentials $e^{jk\omega_0 t}$ are all different for every integer $k$ ; a higher $k$ is always a faster oscillation. A CT periodic signal (e.g. a square wave with sharp jumps) may need infinitely many harmonics, so the synthesis sum runs from $k = -\infty$ to $\infty$ , and convergence must be examined (Dirichlet conditions).
Discrete time: $e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n}e^{j2\pi n} = e^{jk(2\pi/N)n}$ , because $n$ is an integer. Only $N$ distinct exponentials exist, so both the synthesis and analysis sums need only one period ( $N$ terms) and the coefficients repeat, $a_{k+N} = a_k$ . The DTFS is a finite sum, so it always exists and has no convergence problem.

In both cases the analysis is done over one period of the signal; the "infinite" range in CT refers to the infinite number of harmonic terms needed.

2077 Chaitra · 2+2+2 marks

Obtain fourier series coefficients of the signal x[n] = 1 + 3cos(2πn/N + π/3), where N is the fundamental time period of x[n]. Also, plot magnitude and phase spectrum of the coefficients.

Answer

With fundamental period $N$ , $\omega_0 = 2\pi/N$ . Expand the cosine using Euler's formula:

\begin{aligned} x[n] &= 1 + 3\cos\left(\frac{2\pi n}{N} + \frac{\pi}{3}\right) \\ &= 1 + \frac{3}{2}e^{j\pi/3}e^{j(2\pi/N)n} + \frac{3}{2}e^{-j\pi/3}e^{-j(2\pi/N)n} \end{aligned}

Fourier series coefficients

Comparing with $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$ :

a_0 = 1, \qquad a_1 = \frac{3}{2}e^{j\pi/3} = 0.75 + j1.299, \qquad a_{-1} = \frac{3}{2}e^{-j\pi/3} = 0.75 - j1.299

All other coefficients in one period are zero, and $a_{k+N} = a_k$ (so $a_{N-1} = a_{-1}$ , $a_{N+1} = a_1$ ).

$k$	$a_k$	$\lvert a_k\rvert$	$\angle a_k$
$-1$	$1.5e^{-j\pi/3}$	1.5	$-60^\circ$
0	1	1	$0^\circ$
1	$1.5e^{j\pi/3}$	1.5	$+60^\circ$

Magnitude spectrum

 |a_k|  (one period shown; repeats every N)
   1.5     |     |
     1     |  |  |
       -o--+--+--+--o--> k
       -2 -1  0  1  2

Phase spectrum

 angle a_k (deg)
    60           |
       -o--+--o--+--o--> k
   -60     |
       -2 -1  0  1  2

The magnitude is even and the phase odd in $k$ (real signal), and both repeat with period $N$ .

2076 Bhadra · 8 marks

Derive the expression to compute Fourier series coefficients of exponential Fourier series. Explain Gibbs phenomenon.

Answer

Exponential Fourier series coefficients

A periodic signal $x(t)$ with period $T$ and $\omega_0 = 2\pi/T$ is written as

x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t} \qquad \text{(synthesis)}

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period $T$ :

\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

For $k \ne n$ :

\int_0^T e^{j(k-n)\omega_0 t}dt = \frac{e^{j(k-n)\omega_0 T} - 1}{j(k-n)\omega_0} = \frac{e^{j2\pi(k-n)} - 1}{j(k-n)\omega_0} = 0

For $k = n$ the integral is $\int_0^T 1\,dt = T$ . Therefore

\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T \quad\Rightarrow\quad a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \qquad \text{(analysis)}

$a_0 = \frac{1}{T}\int_T x(t)dt$ is the dc value.
$a_k$ is complex: $|a_k|$ is the magnitude spectrum and $\angle a_k$ the phase spectrum.
For real $x(t)$ , $a_{-k} = a_k^*$ .

Example: a square wave of height 1 for $|t| < T_1$ in period $T$ gives $a_0 = 2T_1/T$ and $a_k = \frac{\sin(k\omega_0 T_1)}{k\pi}$ .

Gibbs phenomenon

When a signal with a jump discontinuity (e.g. a square wave) is approximated by a finite number of Fourier terms,

x_N(t) = \sum_{k=-N}^{N}a_k e^{jk\omega_0 t},

the partial sum shows ripples near the discontinuity, with an overshoot of about 9% of the jump height on each side of it.

Key features:

As $N$ increases, the ripples become narrower and crowd closer to the discontinuity, but the peak overshoot does not decrease; it stays at about 9% of the jump (about 1.09 times for a unit jump).
At the discontinuity itself, the series converges to the midpoint $\frac{1}{2}[x(t^-) + x(t^+)]$ .
The energy of the error $\int_T|x(t) - x_N(t)|^2dt \to 0$ as $N \to \infty$ (mean-square convergence), because the area under the ripples shrinks.
It was explained by J. W. Gibbs (1899) after Michelson noticed the overshoot in a mechanical harmonic analyser.

 x_N(t) near a jump (square wave)
       overshoot ~9%
        _/\_/\/\/\________
       |
       |   finite N ripples
 ______|
  \/\/\
       ^ jump at t = T1

Practical meaning: truncating a Fourier series (or ideal filtering, which cuts off high harmonics) produces ringing near sharp edges; window functions are used to reduce it.

2076 Bhadra · 5 marks

A periodic square wave as shown in figure below is defined over one period as: x(t) = A for |t| < T₁; 0 for T₁ < |t| < T/2. Determine the Fourier series coefficients for x(t). [Figure: periodic rectangular pulse train of height A and period T; each pulse is centred on t = 0, ±T, … and extends from −T₁ to T₁]

Answer

The signal has period $T$ , so $\omega_0 = 2\pi/T$ . Over one period ( $-T/2$ to $T/2$ ) it equals $A$ only for $|t| < T_1$ .

a_k = \frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_{-T_1}^{T_1}A\,e^{-jk\omega_0 t}dt

DC term ( $k = 0$ )

a_0 = \frac{1}{T}\int_{-T_1}^{T_1}A\,dt = \frac{2AT_1}{T}

This is the average value (height times duty cycle).

For $k \ne 0$

\begin{aligned} a_k &= \frac{A}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} = \frac{A}{T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{jk\omega_0} \\ &= \frac{2A}{k\omega_0 T}\sin(k\omega_0 T_1) \\ &= \frac{A\sin(k\omega_0 T_1)}{k\pi} \qquad (\omega_0 T = 2\pi) \end{aligned}

Result

a_k = \begin{cases} \dfrac{2AT_1}{T}, & k = 0 \\[2mm] \dfrac{A\sin(k\omega_0 T_1)}{k\pi} = \dfrac{2AT_1}{T}\,\text{sinc}\left(\dfrac{2kT_1}{T}\right), & k \ne 0 \end{cases}

(with $\text{sinc}(x) = \sin(\pi x)/(\pi x)$ ). The coefficients are real and even because $x(t)$ is real and even.

Special case, 50% duty cycle ( $T = 4T_1$ ): $\omega_0 T_1 = \pi/2$ , so $a_0 = A/2$ and $a_k = \frac{A\sin(k\pi/2)}{k\pi}$ : $a_{\pm1} = A/\pi$ , $a_{\pm2} = 0$ , $a_{\pm3} = -A/3\pi$ , $a_{\pm5} = A/5\pi$ , i.e. even harmonics are zero.

 a_k / A  (T = 4T1)
   0.5                 |
 0.318              |  |  |
0.0637  |           |  |  |           |
       -+--o--+--o--+--+--+--o--+--o--+--> k
-0.106        |                 |
       -5 -4 -3 -2 -1  0  1  2  3  4  5

The envelope of $a_k$ follows a sinc shape; making the pulse narrower (smaller $T_1/T$ ) spreads the spectrum wider.

2075 Bhadra · 2+5 marks

State Parseval's relation for discrete time Fourier series. Find the average power of discrete time periodic signal x[n] = sin(πn/3) using Fourier series coefficient.

Answer

Parseval's relation for the DTFS

For a periodic $x[n]$ with period $N$ and coefficients $a_k$ :

\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2

The average power over one period equals the sum of the powers $|a_k|^2$ of the $N$ harmonic components.

Average power of $x[n] = \sin(\pi n/3)$

Period: $\omega_0 = \pi/3$ , $\frac{\omega_0}{2\pi} = \frac{1}{6}$ , so $N = 6$ and the fundamental is $2\pi/6 = \pi/3$ .

Coefficients:

x[n] = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n}

a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = a_5 = -\frac{1}{2j} = \frac{j}{2}

and $a_0 = a_2 = a_3 = a_4 = 0$ in one period.

Power by Parseval:

P = \sum_{k=0}^{5}|a_k|^2 = |a_1|^2 + |a_5|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2}

Check in the time domain: $x[n]$ for $n = 0 \dots 5$ is $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$ :

P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = \frac{3}{6} = 0.5

Answer: average power $P = 0.5$ W.

2075 Bhadra · 6 marks

Show that fourier series coefficient of discrete time signal is periodic in nature.

Answer

For a discrete-time periodic signal $x[n]$ with period $N$ , the DTFS coefficients are

a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}

To show: $a_{k+N} = a_k$

Replace $k$ by $k + N$ :

\begin{aligned} a_{k+N} &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k+N)(2\pi/N)n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}\,e^{-j2\pi n} \end{aligned}

Since $n$ is an integer, $e^{-j2\pi n} = \cos(2\pi n) - j\sin(2\pi n) = 1$ . Hence

a_{k+N} = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n} = a_k

Repeating the argument, $a_{k+mN} = a_k$ for any integer $m$ . So the sequence of DTFS coefficients is periodic in $k$ with period $N$ .

Reason

The root cause is that discrete-time complex exponentials whose frequencies differ by $2\pi$ are identical:

\phi_{k+N}[n] = e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n} = \phi_k[n]

So there are only $N$ distinct harmonics, and the " $(k+N)$ th harmonic" is the same signal as the $k$ th; its coefficient must be the same. This is why the synthesis equation sums over only $N$ consecutive $k$ :

x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}

This is different from the CT Fourier series, where every $e^{jk\omega_0 t}$ is distinct and the coefficients are not periodic.

Example

$x[n] = \cos(\pi n/2)$ , $N = 4$ : $x[n] = \frac{1}{2}e^{j(2\pi/4)n} + \frac{1}{2}e^{-j(2\pi/4)n}$ , so $a_1 = a_{-1} = \frac12$ . Computing from the analysis equation with samples $\{1, 0, -1, 0\}$ :

$k$	$-3$	$-2$	$-1$	0	1	2	3	4	5
$a_k$	0.5	0	0.5	0	0.5	0	0.5	0	0.5

The pattern $\{0, 0.5, 0, 0.5\}$ repeats every 4 values of $k$ , e.g. $a_5 = a_1$ and $a_3 = a_{-1}$ .

 a_k
   0.5  |     |     |     |     |
       -+--o--+--o--+--o--+--o--+--> k
       -3 -2 -1  0  1  2  3  4  5

2075 Baisakh · 5 marks

Find Fourier series coefficients and associated phase of the signal with fundamental frequency ω₀: x(t) = 1 + sin(ω₀t) + 2cos(ω₀t) + cos(2ω₀t + π/4)

Answer

Write each term using Euler's formula, with fundamental frequency $\omega_0$ :

\begin{aligned} x(t) &= 1 + \frac{1}{2j}\left[e^{j\omega_0 t} - e^{-j\omega_0 t}\right] + \left[e^{j\omega_0 t} + e^{-j\omega_0 t}\right] + \frac{1}{2}\left[e^{j(2\omega_0 t + \pi/4)} + e^{-j(2\omega_0 t + \pi/4)}\right] \\ &= 1 + \left(1 + \frac{1}{2j}\right)e^{j\omega_0 t} + \left(1 - \frac{1}{2j}\right)e^{-j\omega_0 t} + \frac{1}{2}e^{j\pi/4}e^{j2\omega_0 t} + \frac{1}{2}e^{-j\pi/4}e^{-j2\omega_0 t} \end{aligned}

Comparing with $x(t) = \sum_k a_k e^{jk\omega_0 t}$ and using $\frac{1}{2j} = -\frac{j}{2}$ :

a_0 = 1, \quad a_1 = 1 - \frac{j}{2}, \quad a_{-1} = 1 + \frac{j}{2}, \quad a_2 = \frac{1}{2}e^{j\pi/4} = \frac{\sqrt2}{4}(1 + j), \quad a_{-2} = \frac{1}{2}e^{-j\pi/4}

$a_k = 0$ for $|k| > 2$ .

Magnitude and phase

|a_{\pm1}| = \sqrt{1^2 + 0.5^2} = \frac{\sqrt5}{2} = 1.118, \qquad \angle a_{\pm1} = \mp\tan^{-1}(0.5) = \mp 26.57^\circ

$k$	$a_k$	$\lvert a_k\rvert$	$\angle a_k$
$-2$	$0.354 - j0.354$	0.5	$-45^\circ$ ( $-\pi/4$ )
$-1$	$1 + j0.5$	1.118	$+26.57^\circ$ (0.464 rad)
0	1	1	$0^\circ$
1	$1 - j0.5$	1.118	$-26.57^\circ$ ( $-0.464$ rad)
2	$0.354 + j0.354$	0.5	$+45^\circ$ ( $\pi/4$ )

 |a_k|
 1.118     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

 angle a_k (deg)
    45              |
 26.57     |        |
       -+--+--o--+--+--> k
-26.57  |        |
   -45  |
       -2 -1  0  1  2

As expected for a real signal, $a_{-k} = a_k^*$ : the magnitude is even and the phase is odd.

2074 Bhadra · 3 marks

Explain the Dirichlet conditions for convergence of Fourier series.

Answer

The Dirichlet conditions are a set of sufficient conditions on a periodic signal $x(t)$ (period $T$ ) which guarantee that its Fourier series converges to $x(t)$ at every point of continuity.

Absolute integrability over one period: the signal must be absolutely integrable over a period,

\int_{T} |x(t)|\, dt < \infty

This makes every coefficient finite, since $|a_k| \le \frac{1}{T}\int_T |x(t)|\,dt$ .

Violating example: $x(t) = 1/t$ for $0 < t \le 1$ , repeated with period 1.

Finite number of maxima and minima: in any one period, $x(t)$ $x (t)$ has only a finite number of maxima and minima (bounded variation).
- Violating example: $x(t) = \sin(2\pi/t)$ for $0 < t \le 1$ .
Finite number of discontinuities: in any finite interval there are only a finite number of discontinuities, and each one is finite (a finite jump).
- Violating example: a staircase signal whose steps get halved again and again within one period, giving infinitely many jumps.

Result: if these conditions hold, the Fourier series equals $x(t)$ wherever $x(t)$ is continuous. At a jump it converges to the average of the left and right limits, $\frac{1}{2}[x(t^-) + x(t^+)]$ . Near a jump there is still an overshoot of about 9% (the Gibbs phenomenon).

The conditions are sufficient but not necessary. Signals that break them are mostly artificial, so almost every practical signal has a convergent Fourier series.

2074 Bhadra · 4 marks

Find Fourier series coefficient of discrete time periodic signal x[n] = sin ω₀n where ω₀ = 2π/N. Plot the signal when N = 5.

Answer

Given $x[n] = \sin \omega_0 n$ with $\omega_0 = 2\pi/N$ , the signal is periodic with fundamental period $N$ . Its DTFS coefficients can be found by inspection, using Euler's formula.

Fourier series coefficients

x[n] = \sin\left(\frac{2\pi}{N}n\right) = \frac{1}{2j}e^{j(2\pi/N)n} - \frac{1}{2j}e^{-j(2\pi/N)n}

Compare this with the synthesis equation $x[n] = \sum_{k=\langle N\rangle} a_k e^{jk(2\pi/N)n}$ :

a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2}

All other coefficients in one period are zero. DTFS coefficients repeat with period $N$ , so $a_{k+N} = a_k$ :

$a_1 = a_{1+N} = a_{1-N} = \dots = -j/2$
$a_{-1} = a_{N-1} = a_{2N-1} = \dots = j/2$

Over $k = 0, 1, \dots, N-1$ : $a_1 = -j/2$ , $a_{N-1} = j/2$ , and every other $a_k = 0$ .

For N = 5: $a_1 = -j/2$ , $a_4 = j/2$ , and $a_0 = a_2 = a_3 = 0$ , repeating every 5 values of $k$ . In magnitude form $|a_1| = |a_4| = 1/2$ , with $\angle a_1 = -\pi/2$ and $\angle a_4 = +\pi/2$ .

Plot of the signal for N = 5

$x[n] = \sin(2\pi n/5)$ :

n	0	1	2	3	4
x[n]	0	0.951	0.588	−0.588	−0.951

 x[n]
  0.95 |     *              *
  0.59 |     |  *           |  *
       |     |  |           |  |
  0  --*-----+--+--+--+--*--+--+--+--+-- n
       0     1  2  3  4  5  6  7  8  9
 -0.59 |           |  |           |  |
 -0.95 |           *  *           *  *

The pattern repeats every 5 samples.

2073 Magh · 6 marks

Derive the expression of continuous time exponential Fourier series x[n] = A sin ω₀n

Answer

The signal is read as the continuous-time sinusoid $x(t) = A\sin\omega_0 t$ (the question writes $x[n]$ , but asks for the continuous-time series). The derivation of the general coefficient formula comes first, then the formula is applied.

Exponential Fourier series and its coefficients

A periodic signal with period $T$ and $\omega_0 = 2\pi/T$ can be written as a sum of harmonically related complex exponentials:

x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}

To find $a_n$ , multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty} a_k \int_0^T e^{j(k-n)\omega_0 t}dt

Orthogonality of the exponentials gives

\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}

so only the $k = n$ term remains, and

a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt

Applying it to x(t) = A sin ω₀t

Using Euler's identity:

x(t) = A\,\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} = \frac{A}{2j}e^{j\omega_0 t} - \frac{A}{2j}e^{-j\omega_0 t}

Comparing with $\sum a_k e^{jk\omega_0 t}$ :

a_1 = \frac{A}{2j} = -j\frac{A}{2}, \qquad a_{-1} = -\frac{A}{2j} = j\frac{A}{2}, \qquad a_k = 0 \ (k \ne \pm 1)

The integral formula gives the same result. For example,

a_1 = \frac{1}{T}\int_0^T A\,\frac{e^{j\omega_0 t}-e^{-j\omega_0 t}}{2j}e^{-j\omega_0 t}dt = \frac{A}{2j}\cdot\frac{1}{T}\left[T - 0\right] = \frac{A}{2j}

Spectrum

Magnitude: $|a_1| = |a_{-1}| = A/2$
Phase: $\angle a_1 = -\pi/2$ , $\angle a_{-1} = +\pi/2$

 |a_k|               angle(a_k)
  A/2  |     |        +pi/2 |
       |     |              |
 ------+--+--+--  k   --+---+---+--  k
      -1  0  1         -1   0   1
                             |
                       -pi/2 |

The spectrum has only two lines, at $\pm\omega_0$ . The DT signal $A\sin\omega_0 n$ with $\omega_0 = 2\pi/N$ gives the same coefficients $a_{\pm1} = \pm A/(2j)$ , repeated every $N$ .

2073 Magh · 5 marks

Find Fourier series coefficients of signal and plot the coefficients. [The signal is not printed on the paper.]

Answer

The paper does not print the signal. This answer takes the standard textbook case: a periodic square wave of period $T$ with $x(t) = 1$ for $|t| < T_1$ and $x(t) = 0$ for $T_1 < |t| < T/2$ .

 x(t)
  1     ___       ___       ___
       |   |     |   |     |   |
 ______|   |_____|   |_____|   |___ t
     -T1   T1  T-T1  T+T1
       (pulse centred at t = 0)

Coefficients

With $\omega_0 = 2\pi/T$ , the DC term is

a_0 = \frac{1}{T}\int_{-T_1}^{T_1} 1\,dt = \frac{2T_1}{T}

For $k \ne 0$ :

\begin{aligned} a_k &= \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}dt = \frac{1}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} \\ &= \frac{2}{k\omega_0 T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{2j} = \frac{\sin(k\omega_0 T_1)}{k\pi} \end{aligned}

The coefficients are real, because $x(t)$ is real and even.

Example values for T = 4T₁ (50% duty cycle)

Here $\omega_0 T_1 = \pi/2$ , so $a_k = \frac{\sin(k\pi/2)}{k\pi}$ :

k	0	±1	±2	±3	±4	±5
$a_k$	1/2	$1/\pi$ ≈ 0.318	0	$-1/(3\pi)$ ≈ −0.106	0	$1/(5\pi)$ ≈ 0.064

Plot of the coefficients

 a_k
  0.50                   |
  0.32                |  |  |
  0.06    |           |  |  |           |
  0  -----+--+--+--+--+--+--+--+--+--+--+--- k
         -5 -4 -3 -2 -1  0  1  2  3  4  5
 -0.11          |                 |

The coefficients are samples, at $\omega = k\omega_0$ , of the sinc-shaped envelope $\frac{\omega_0}{\pi}\cdot\frac{\sin(\omega T_1)}{\omega}$ . Even harmonics vanish for a 50% duty cycle, and the odd ones decrease as $1/k$ with alternating sign.

2073 Magh · 3+3 marks

State and prove the frequency shifting and convolution properties for discrete time Fourier series pair.

Answer

Let $x[n]$ and $y[n]$ be periodic with the same period $N$ , and $\omega_0 = 2\pi/N$ , with $x[n] \xleftrightarrow{FS} a_k$ and $y[n] \xleftrightarrow{FS} b_k$ , where

a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-jk\omega_0 n}

Frequency shifting property

Statement: multiplying a signal by a harmonic exponential shifts its coefficients:

e^{jM\omega_0 n}x[n] \xleftrightarrow{FS} a_{k-M}

Proof: let $z[n] = e^{jM(2\pi/N)n}x[n]$ . Its coefficients are

\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle} e^{jM\omega_0 n}x[n]e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-j(k-M)\omega_0 n} = a_{k-M} \end{aligned}

So multiplying by $e^{jM\omega_0 n}$ moves the whole spectrum $M$ places to the right. This is the DT form of modulation.

Convolution property (periodic convolution)

Statement: periodic convolution in time corresponds to multiplication of the coefficients, scaled by $N$ :

\sum_{r=\langle N\rangle} x[r]\,y[n-r] \xleftrightarrow{FS} N a_k b_k

Proof: let $z[n] = \sum_{r=\langle N\rangle} x[r]y[n-r]$ , which is also periodic with period $N$ . Then

\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}\sum_{r=\langle N\rangle} x[r]\,y[n-r]\,e^{-jk\omega_0 n} \\ &= \frac{1}{N}\sum_{r=\langle N\rangle} x[r]e^{-jk\omega_0 r}\sum_{n=\langle N\rangle} y[n-r]e^{-jk\omega_0 (n-r)} \end{aligned}

Put $m = n - r$ . Because $y[m]e^{-jk\omega_0 m}$ is periodic with period $N$ , the inner sum over any $N$ consecutive $m$ is

\sum_{m=\langle N\rangle} y[m]e^{-jk\omega_0 m} = N b_k

Therefore

c_k = \frac{1}{N}\cdot N b_k\sum_{r=\langle N\rangle} x[r]e^{-jk\omega_0 r} = N b_k\cdot a_k = N a_k b_k

Dual (multiplication) property: $x[n]y[n] \xleftrightarrow{FS} \sum_{l=\langle N\rangle} a_l b_{k-l}$ , which is periodic convolution of the coefficients.

2072 Magh · 5 marks

Find the Fourier Series coefficient of the following continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2, with fundamental frequency π.

Answer

The fundamental frequency is $\omega_0 = \pi$ , so the period is $T = 2\pi/\omega_0 = 2$ . Over one period, $x(t) = 1.5$ on $0 \le t < 1$ and $x(t) = -1.5$ on $1 \le t < 2$ .

 x(t)
  1.5  ____      ____      ____
      |    |    |    |    |
 -----+----+----+----+----+----- t
      0    1    2    3    4
           |____|    |____|
 -1.5

DC coefficient

a_0 = \frac{1}{T}\int_0^2 x(t)\,dt = \frac{1}{2}\left[1.5(1) - 1.5(1)\right] = 0

Coefficients for k ≠ 0

a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \frac{1}{2}\left[\int_0^1 1.5\,e^{-jk\pi t}dt - \int_1^2 1.5\,e^{-jk\pi t}dt\right]

Evaluate each integral, using $e^{-jk\pi} = (-1)^k$ and $e^{-j2k\pi} = 1$ :

\begin{aligned} \int_0^1 e^{-jk\pi t}dt &= \frac{1 - e^{-jk\pi}}{jk\pi} = \frac{1-(-1)^k}{jk\pi} \\ \int_1^2 e^{-jk\pi t}dt &= \frac{e^{-jk\pi} - e^{-j2k\pi}}{jk\pi} = \frac{(-1)^k - 1}{jk\pi} \end{aligned}

Substituting:

\begin{aligned} a_k &= \frac{1.5}{2}\cdot\frac{[1-(-1)^k] - [(-1)^k - 1]}{jk\pi} \\ &= \frac{1.5}{2}\cdot\frac{2[1-(-1)^k]}{jk\pi} = \frac{1.5\,[1-(-1)^k]}{jk\pi} \end{aligned}

So

a_k = \begin{cases} \dfrac{3}{jk\pi} = -j\dfrac{3}{k\pi}, & k \text{ odd} \\ 0, & k \text{ even} \end{cases}

k	±1	±2	±3	±5
$\lvert a_k\rvert$	$3/\pi$ ≈ 0.955	0	$1/\pi$ ≈ 0.318	$3/(5\pi)$ ≈ 0.191
$\angle a_k$	∓π/2	—	∓π/2	∓π/2

The coefficients are purely imaginary and odd ( $a_{-k} = -a_k$ ). This is expected, because $x(t)$ is real and has odd symmetry about $t = 0$ (half-wave symmetry also removes the even harmonics). In trigonometric form:

x(t) = \frac{6}{\pi}\left[\sin\pi t + \frac{1}{3}\sin 3\pi t + \frac{1}{5}\sin 5\pi t + \dots\right]

Answer: $a_0 = 0$ , $a_k = -j\,3/(k\pi)$ for odd $k$ , and $a_k = 0$ for even $k$ .

2072 Magh · 2+4 marks

State and prove the convolution property of continuous-time Fourier series.

Answer

Statement

Let $x(t)$ and $y(t)$ be periodic with the same period $T$ ( $\omega_0 = 2\pi/T$ ), with $x(t) \xleftrightarrow{FS} a_k$ and $y(t) \xleftrightarrow{FS} b_k$ . The periodic convolution of the two signals has Fourier series coefficients $T a_k b_k$ :

z(t) = \int_T x(\tau)\,y(t-\tau)\,d\tau \;\xleftrightarrow{FS}\; c_k = T a_k b_k

So convolution in time becomes multiplication of the coefficients.

Proof

$z(t)$ is periodic with period $T$ , because $y(t - \tau)$ is periodic in $t$ . Its coefficients are

c_k = \frac{1}{T}\int_T z(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T\left[\int_T x(\tau)y(t-\tau)d\tau\right]e^{-jk\omega_0 t}dt

Swap the order of integration, and write $e^{-jk\omega_0 t} = e^{-jk\omega_0 \tau}e^{-jk\omega_0 (t-\tau)}$ :

c_k = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}\left[\int_T y(t-\tau)e^{-jk\omega_0 (t-\tau)}dt\right]d\tau

Put $\sigma = t - \tau$ . The integrand is periodic with period $T$ , so integrating over any interval of length $T$ gives the same result:

\int_T y(\sigma)e^{-jk\omega_0 \sigma}d\sigma = T b_k

Therefore

c_k = \frac{1}{T}\cdot T b_k\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = T b_k\cdot a_k

\boxed{c_k = T a_k b_k}

Use: the output of an LTI system for a periodic input can be found by multiplying spectra instead of convolving. The dual result is the multiplication property: $x(t)y(t) \xleftrightarrow{FS} \sum_{l} a_l b_{k-l}$ .

2072 Asoj · 6 marks

State and prove the linearity and time shift properties of discrete time Fourier series representation.

Answer

Let $x[n]$ and $y[n]$ be periodic with period $N$ ( $\omega_0 = 2\pi/N$ ), with $x[n] \xleftrightarrow{FS} a_k$ and $y[n] \xleftrightarrow{FS} b_k$ , where

a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-jk\omega_0 n}

Linearity

Statement: for any constants $A$ and $B$ ,

z[n] = A x[n] + B y[n] \xleftrightarrow{FS} c_k = A a_k + B b_k

Proof: $z[n]$ is also periodic with period $N$ . Then

\begin{aligned} c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}\left(A x[n] + B y[n]\right)e^{-jk\omega_0 n} \\ &= A\cdot\frac{1}{N}\sum_{n=\langle N\rangle} x[n]e^{-jk\omega_0 n} + B\cdot\frac{1}{N}\sum_{n=\langle N\rangle} y[n]e^{-jk\omega_0 n} \\ &= A a_k + B b_k \end{aligned}

Example: if $x[n] = \cos(2\pi n/N)$ ( $a_{\pm1} = 1/2$ ) and $y[n] = 1$ ( $b_0 = 1$ ), then $3x[n] + 2y[n]$ has $c_0 = 2$ and $c_{\pm1} = 3/2$ .

Time shifting

Statement: a delay of $n_0$ samples multiplies each coefficient by a linear-phase term:

x[n - n_0] \xleftrightarrow{FS} a_k e^{-jk\omega_0 n_0} = a_k e^{-jk(2\pi/N)n_0}

Proof: let $z[n] = x[n - n_0]$ . Its coefficients are

c_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n-n_0]e^{-jk\omega_0 n}

Put $m = n - n_0$ , so $n = m + n_0$ . As $n$ runs over one period, so does $m$ :

\begin{aligned} c_k &= \frac{1}{N}\sum_{m=\langle N\rangle} x[m]e^{-jk\omega_0 (m+n_0)} \\ &= e^{-jk\omega_0 n_0}\cdot\frac{1}{N}\sum_{m=\langle N\rangle} x[m]e^{-jk\omega_0 m} = e^{-jk\omega_0 n_0}a_k \end{aligned}

Meaning: $|c_k| = |a_k|$ , so a time shift does not change the magnitude spectrum. It only adds a phase of $-k\omega_0 n_0$ , which is linear in $k$ .

2071 Magh · 5+1 marks

Find Fourier series coefficients for periodic rectangular pulses with unity amplitude. Draw the magnitude spectrum.

Answer

Take a periodic train of rectangular pulses with unity amplitude, period $T$ and pulse width $2T_1$ , centred at $t = 0$ :

x(t) = \begin{cases} 1, & |t| < T_1 \\ 0, & T_1 < |t| < T/2 \end{cases} \qquad x(t+T) = x(t), \quad \omega_0 = \frac{2\pi}{T}

 x(t)
  1     ___       ___       ___
       |   |     |   |     |   |
 ______|   |_____|   |_____|   |___ t
     -T1   T1  T-T1  T+T1
       (width 2T1, period T)

Fourier series coefficients

DC term (the average value):

a_0 = \frac{1}{T}\int_{-T_1}^{T_1} dt = \frac{2T_1}{T}

For $k \ne 0$ :

\begin{aligned} a_k &= \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}dt = \frac{1}{T}\cdot\frac{e^{-jk\omega_0 T_1} - e^{jk\omega_0 T_1}}{-jk\omega_0} \\ &= \frac{2}{k\omega_0 T}\sin(k\omega_0 T_1) = \frac{\sin(k\omega_0 T_1)}{k\pi} \end{aligned}

This can also be written $a_k = \frac{2T_1}{T}\,\mathrm{sinc}\!\left(\frac{2kT_1}{T}\right)$ , with $\mathrm{sinc}(x) = \frac{\sin \pi x}{\pi x}$ .

Example: T = 4T₁ (duty cycle 50%)

$a_k = \frac{\sin(k\pi/2)}{k\pi}$ : $a_0 = 0.5$ , $a_{\pm1} = 1/\pi \approx 0.318$ , $a_{\pm2} = 0$ , $a_{\pm3} = -1/(3\pi) \approx -0.106$ , $a_{\pm5} \approx 0.064$ .

Magnitude spectrum

 |a_k|
 0.5              |
 0.32          |  |  |
 0.11       |  |  |  |  |
 0.06    |  |  |  |  |  |  |
 --------+--+--+--+--+--+--+--- k
        -5 -3 -1  0  1  3  5
     (|a_k| = 0 for even k != 0)

The lines sit at $k\omega_0$ under a sinc-shaped envelope. A wider pulse ( $T_1$ larger) gives a narrower envelope. A longer period ( $T$ larger) gives more closely spaced lines, and as $T \to \infty$ they merge into the continuous Fourier transform $2\sin(\omega T_1)/\omega$ .

2071 Bhadra · 6 marks

Discuss the following properties of continuous time Fourier series. (a) Time shifting (b) Time scaling (c) Conjugation

Answer

Let $x(t)$ be periodic with period $T$ ( $\omega_0 = 2\pi/T$ ) and $x(t) \xleftrightarrow{FS} a_k$ , where $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ .

(a) Time shifting

x(t - t_0) \xleftrightarrow{FS} a_k e^{-jk\omega_0 t_0}

Proof: put $\tau = t - t_0$ :

b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt = e^{-jk\omega_0 t_0}\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = e^{-jk\omega_0 t_0}a_k

The magnitudes $|b_k| = |a_k|$ stay the same. Only a linear phase $-k\omega_0 t_0$ is added.

(b) Time scaling

If $\alpha > 0$ , then $x(\alpha t)$ is periodic with period $T/\alpha$ and fundamental frequency $\alpha\omega_0$ . Its Fourier series is

x(\alpha t) = \sum_{k=-\infty}^{\infty} a_k e^{jk(\alpha\omega_0)t}

Proof: with period $T' = T/\alpha$ , put $\tau = \alpha t$ :

b_k = \frac{\alpha}{T}\int_{T/\alpha} x(\alpha t)e^{-jk(\alpha\omega_0)t}dt = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0\tau}d\tau = a_k

The coefficients do not change. What changes is the spacing of the harmonics, which becomes $\alpha\omega_0$ instead of $\omega_0$ . Compressing a signal in time spreads its spectral lines further apart.

(c) Conjugation

x^*(t) \xleftrightarrow{FS} a_{-k}^*

Proof: take the conjugate of the synthesis equation:

x^*(t) = \sum_{k} a_k^* e^{-jk\omega_0 t} = \sum_{k} a_{-k}^* e^{jk\omega_0 t} \quad (k \to -k)

So the coefficients of $x^*(t)$ are $a_{-k}^*$ .

Consequence (conjugate symmetry): if $x(t)$ is real, then $x(t) = x^*(t)$ , so $a_k = a_{-k}^*$ . This gives $|a_k| = |a_{-k}|$ (even magnitude) and $\angle a_{-k} = -\angle a_k$ (odd phase). If $x(t)$ is also even, the $a_k$ are real and even.

2070 Magh · 4+4 marks

How could you represent a signal x(t) with harmonically related exponentials? State and prove conjugation and conjugate symmetry property of CTFS.

Answer

Representation with harmonically related exponentials

Take a periodic signal $x(t)$ with fundamental period $T$ and $\omega_0 = 2\pi/T$ . The set of harmonically related complex exponentials

\phi_k(t) = e^{jk\omega_0 t}, \quad k = 0, \pm1, \pm2, \dots

are all periodic with period $T$ , since each frequency is an integer multiple of $\omega_0$ . The term for $k = 0$ is the DC part, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$ th harmonics. A linear combination of them is periodic with period $T$ :

x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t} \quad \text{(synthesis equation)}

The functions are orthogonal over one period: $\int_T e^{jk\omega_0 t}e^{-jn\omega_0 t}dt$ equals $T$ when $k = n$ and 0 otherwise. Multiplying the synthesis equation by $e^{-jn\omega_0 t}$ and integrating over $T$ therefore leaves only one term. This gives

a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \quad \text{(analysis equation)}

The series converges to $x(t)$ when $x(t)$ meets the Dirichlet conditions. Example: $\cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$ , so $a_{\pm1} = 1/2$ .

Conjugation property

Statement: if $x(t) \xleftrightarrow{FS} a_k$ , then

x^*(t) \xleftrightarrow{FS} a_{-k}^*

Proof: let $b_k$ be the coefficients of $x^*(t)$ :

b_k = \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{jk\omega_0 t}dt\right]^* = \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^*

Conjugate symmetry property

Statement: if $x(t)$ is real, its coefficients are conjugate symmetric:

a_{-k} = a_k^*

Proof: for a real signal $x(t) = x^*(t)$ , so the two signals have the same coefficients: $a_k = b_k = a_{-k}^*$ . Taking the conjugate of both sides gives $a_k^* = a_{-k}$ .

Consequences for real $x(t)$ :

$|a_{-k}| = |a_k|$ , so the magnitude spectrum is even.
$\angle a_{-k} = -\angle a_k$ , so the phase spectrum is odd.
$\mathrm{Re}\{a_k\}$ is even and $\mathrm{Im}\{a_k\}$ is odd.
Real and even $x(t)$ gives real, even $a_k$ . Real and odd $x(t)$ gives purely imaginary, odd $a_k$ .

For example, $\sin\omega_0 t$ has $a_1 = 1/(2j)$ and $a_{-1} = -1/(2j) = a_1^*$ .

2070 Bhadra · 1+6 marks

What is the information provided by Fourier series coefficients of a signal? State and prove time shifting and conjugation properties of continuous time Fourier series representation.

Answer

Information provided by Fourier series coefficients

The coefficient $a_k$ gives the amount (magnitude $|a_k|$ ) and the phase ( $\angle a_k$ ) of the $k$ th harmonic $e^{jk\omega_0 t}$ in the signal. This makes the coefficients the frequency-domain picture of the signal. $a_0$ is the DC (average) value, $|a_k|^2$ is the power at frequency $k\omega_0$ (by Parseval, $P = \sum|a_k|^2$ ), and the spread of the $a_k$ shows the bandwidth of the signal.

Time shifting property

Statement: if $x(t) \xleftrightarrow{FS} a_k$ with period $T$ and $\omega_0 = 2\pi/T$ , then

x(t - t_0) \xleftrightarrow{FS} e^{-jk\omega_0 t_0}a_k

Proof: the coefficients of $y(t) = x(t - t_0)$ are

b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt

Put $\tau = t - t_0$ (so $dt = d\tau$ ). The range of $\tau$ is still one full period:

\begin{aligned} b_k &= \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0(\tau + t_0)}d\tau \\ &= e^{-jk\omega_0 t_0}\cdot\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau = e^{-jk\omega_0 t_0}a_k \end{aligned}

Interpretation: $|b_k| = |a_k|$ , so the magnitude spectrum does not change. The phase changes by $-k\omega_0 t_0$ , which is proportional to the harmonic number.

Conjugation property

Statement:

x^*(t) \xleftrightarrow{FS} a_{-k}^*

Proof: the coefficients of $x^*(t)$ are

\begin{aligned} b_k &= \frac{1}{T}\int_T x^*(t)e^{-jk\omega_0 t}dt = \left[\frac{1}{T}\int_T x(t)e^{+jk\omega_0 t}dt\right]^* \\ &= \left[\frac{1}{T}\int_T x(t)e^{-j(-k)\omega_0 t}dt\right]^* = a_{-k}^* \end{aligned}

Consequence: if $x(t)$ is real, then $x(t) = x^*(t)$ , which gives $a_{-k} = a_k^*$ (conjugate symmetry). The magnitude spectrum is then even and the phase spectrum is odd.

2070 Bhadra · 5+2 marks

Find out Fourier series coefficients of a periodic discrete time signal described over a period as x[n] = 2 for |n| ≤ 1; 0 for 1 < |n| ≤ 3. Using the Fourier series coefficients calculated above, find the Fourier series coefficients of the signal e^(j4πn/7) x[n].

Answer

The signal is $x[n] = 2$ for $n = -1, 0, 1$ and $x[n] = 0$ for $n = \pm2, \pm3$ . One period runs over $n = -3$ to $3$ , so $N = 7$ and $\omega_0 = 2\pi/7$ .

 x[n]
  2        *  *  *              *  *  *
           |  |  |              |  |  |
--*--*--*--+--+--+--*--*--*--*--+--+--+--- n
 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8

Fourier series coefficients of x[n]

a_k = \frac{1}{N}\sum_{n=-1}^{1} 2\,e^{-jk(2\pi/7)n} = \frac{2}{7}\left[1 + e^{-j2\pi k/7} + e^{j2\pi k/7}\right] = \frac{2}{7}\left[1 + 2\cos\frac{2\pi k}{7}\right]

In closed form (the standard result for a DT rectangular pulse with $N_1 = 1$ ):

a_k = \frac{2}{7}\cdot\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}, \quad k \ne 0, \pm7, \dots; \qquad a_0 = \frac{2(2N_1+1)}{N} = \frac{6}{7}

k	0	1	2	3	4	5	6
$a_k$	0.857	0.642	0.159	−0.229	−0.229	0.159	0.642

The coefficients are real and even ( $a_k = a_{-k} = a_{7-k}$ ), because $x[n]$ is real and even. They repeat with period 7.

Coefficients of y[n] = e^{j4πn/7} x[n]

e^{j4\pi n/7} = e^{j2(2\pi/7)n} = e^{jM\omega_0 n} \quad \text{with } M = 2

By the frequency shifting property, $e^{jM\omega_0 n}x[n] \xleftrightarrow{FS} a_{k-M}$ , so

b_k = a_{k-2} = \frac{2}{7}\cdot\frac{\sin\left(3\pi(k-2)/7\right)}{\sin\left(\pi(k-2)/7\right)}, \qquad b_2 = \frac{6}{7}

k	0	1	2	3	4	5	6
$b_k = a_{k-2}$	0.159	0.642	0.857	0.642	0.159	−0.229	−0.229

Answer: $a_k = \frac{2}{7}\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}$ with $a_0 = 6/7$ , and $b_k = a_{k-2}$ . The spectrum is the same shape, moved 2 places to the right (peak at $k = 2$ ).

2069 Bhadra · 6 marks

Find the Fourier series representation of the signal x[n] = Σ_{ℓ=−∞}^{∞} δ[n − ℓN].

Answer

The signal is a periodic impulse train: a unit impulse at every multiple of $N$ .

x[n] = \sum_{\ell=-\infty}^{\infty}\delta[n - \ell N]

 x[n]
  1  *              *              *
     |              |              |
 ----+--*--*--...---+--*--*--...---+---- n
     0              N              2N

It is periodic with period $N$ , so $\omega_0 = 2\pi/N$ .

Coefficients

a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]\,e^{-jk(2\pi/N)n}

Choose the period $n = 0, 1, \dots, N-1$ . Within it $x[n] = \delta[n]$ , which is 1 at $n = 0$ and 0 elsewhere:

a_k = \frac{1}{N}\sum_{n=0}^{N-1}\delta[n]\,e^{-jk(2\pi/N)n} = \frac{1}{N}e^{0} = \frac{1}{N} \quad \text{for all } k

Fourier series representation

x[n] = \sum_{k=0}^{N-1}\frac{1}{N}e^{jk(2\pi/N)n} = \frac{1}{N}\sum_{k=\langle N\rangle} e^{jk(2\pi/N)n}

Check: for $n = 0, \pm N, \dots$ every exponential equals 1, so the sum is $N\cdot\frac{1}{N} = 1$ . For other $n$ the sum is a full geometric series over the $N$ th roots of unity, which is $\frac{1}{N}\cdot\frac{1 - e^{j2\pi n}}{1 - e^{j2\pi n/N}} = 0$ . This matches $x[n]$ .

Spectrum: all $N$ harmonics have the same amplitude $1/N$ and zero phase. The spectrum is flat. A train of narrow impulses contains all frequencies equally, just as the CT impulse train $\sum\delta(t - kT)$ has $a_k = 1/T$ . This result is the basis of the sampling theorem.

2083 Baisakh (new course) · 4 marks

Describe the continuous-time Fourier series' time reversal, time scaling, and frequency shifting properties.

Answer

Let $x(t)$ be periodic with period $T$ , $\omega_0 = 2\pi/T$ , and $x(t) \xleftrightarrow{FS} a_k$ .

Time reversal

x(-t) \xleftrightarrow{FS} a_{-k}

Put $k = -m$ in the synthesis equation:

x(-t) = \sum_k a_k e^{-jk\omega_0 t} = \sum_m a_{-m}e^{jm\omega_0 t}

Reversing the signal in time reverses its coefficient sequence. If $x(t)$ is even, $a_{-k} = a_k$ . If it is odd, $a_{-k} = -a_k$ .

Time scaling

x(\alpha t),\ \alpha > 0: \quad x(\alpha t) = \sum_k a_k e^{jk(\alpha\omega_0)t}

$x(\alpha t)$ has period $T/\alpha$ and fundamental frequency $\alpha\omega_0$ . The coefficients $a_k$ stay the same, and only the harmonic spacing changes from $\omega_0$ to $\alpha\omega_0$ . Compressing the signal ( $\alpha > 1$ ) spreads the spectral lines apart.

Frequency shifting

e^{jM\omega_0 t}x(t) \xleftrightarrow{FS} a_{k-M}

The coefficients of the product are

\frac{1}{T}\int_T e^{jM\omega_0 t}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(t)e^{-j(k-M)\omega_0 t}dt = a_{k-M}

Multiplying by a harmonic exponential moves the whole line spectrum $M$ places to the right. This is the dual of the time-shift property.

Past questions and answers

1.Derive the exponential form of the Fourier series of a continuous-time periodic signal x(t) (synthesis and analysis equations).

Synthesis equation

Analysis equation

The Fourier series pair

2.State and prove Parseval's relation for Discrete Time Fourier Series (discrete time periodic signals).

Proof

Meaning

3.Derive the exponential form of the Fourier series (synthesis and analysis equations) for a discrete time periodic signal x[n] with period N.

Only NNN distinct exponentials

Synthesis equation

Orthogonality result

Analysis equation

The DTFS pair

4.What is Parseval's relation? State and prove Parseval's relation for continuous-time periodic signals.

Statement (continuous-time periodic signals)

Proof

Interpretation

5.Derive the trigonometric Fourier series expansion of a continuous-time periodic signal x(t).

Orthogonality relations (over one period TTT)

Finding a0a_0a0​

Finding ana_nan​

Finding bnb_nbn​

Summary and notes

6.Find and plot the Fourier series coefficients (magnitude and associated phase) for the following signal with period N. x[n] = 1 + sin((2π/N)n) + 3cos((2π/N)n) + cos((4π/N)n + π/2)

Fourier series coefficients (one period, k=−2k = -2k=−2 to 222)

Plots (one period, k=−2…2k = -2 \dots 2k=−2…2, repeating every NNN)

7.State and prove the time-scaling property for continuous-time (exponential) Fourier series pair.

Statement

Proof

Interpretation

8.Consider a signal x[n] = sin(ω₀n) with period N = 6. Find the Fourier series coefficient of the given signal x[n]. Plot spectra. Also find the average power.

Fourier series coefficients

Spectra

Average power

9.State and prove the time-scaling property for discrete-time Fourier series pair.

Statement

Proof

Interpretation

10.Find the frequency domain representation (both magnitude and phase plot) of a periodic signal defined as x[n] = {1, −0.5, 0, 0, 0.5} (x[0] = 1), with period N = 5.

Values over one period

Magnitude and phase plots (repeat every 5)

11.Find the Fourier series coefficients of the signal x[n] = 1 + sin[(2π/N)n] + 3cos[(2π/N)n] + cos[(4π/N)n]

12.State and prove the time-shifting property for discrete-time Fourier series pair.

Statement

Proof

Interpretation

13.State and prove time shifting and time scaling properties of continuous time Fourier series.

Time shifting property

Time scaling property

Summary

14.Describe the time shifting and frequency shifting properties of continuous time Fourier series. Find Fourier series coefficient of signal x(t) = A cos(ω₀t).

Time shifting property

Frequency shifting property

Fourier series coefficients of x(t)=Acos⁡(ω0t)x(t) = A\cos(\omega_0 t)x(t)=Acos(ω0​t)

15.State and prove frequency shifting property of discrete time Fourier series.

Statement

Proof

16.What is Fourier series representation of a signal? Derive an expression for analysis and synthesis equation of the Fourier Series for a CT periodic signal x(t).

Synthesis equation

Analysis equation

Summary

17.What are the significance of Fourier series coefficients of a signal? State and prove time scaling and conjugation properties of continuous time Fourier series representation.

Significance of Fourier series coefficients

Time scaling property

Conjugation property

18.Calculate and plot magnitude and phase spectrum of the following periodic signal. [Figure: sawtooth wave of period 1: x(t) rises linearly from 0 at t = 0 to 1 at t = 1 and then drops back to 0, i.e. x(t) = t for 0 ≤ t < 1, repeating every 1 s; shown from t = −3 to t = 4]

DC term (k=0k = 0k=0)

Coefficients for k≠0k \ne 0k=0

Magnitude and phase

Spectra (lines at ω=2πk\omega = 2\pi kω=2πk)

19."Even if a signal is periodic, its Fourier series might not exist." Justify the statement using three different examples.

Example 1: not absolutely integrable

Example 2: infinite number of maxima and minima

Example 3: infinite number of discontinuities

20.Derive the formula for synthesis equation and analysis equation of Fourier series. Mention the three conditions that guarantee the existence of Fourier series.

Synthesis equation

Analysis equation

Conditions for existence (Dirichlet conditions)

21.State and explain four properties of discrete time Fourier series.

Derive the exponential form of the Fourier series of a continuous-time periodic signal x(t) (synthesis and analysis equations).

State and prove Parseval's relation for Discrete Time Fourier Series (discrete time periodic signals).

Derive the exponential form of the Fourier series (synthesis and analysis equations) for a discrete time periodic signal x[n] with period N.

Only $N$ distinct exponentials

What is Parseval's relation? State and prove Parseval's relation for continuous-time periodic signals.

Derive the trigonometric Fourier series expansion of a continuous-time periodic signal x(t).

Orthogonality relations (over one period $T$ )

Finding $a_0$

Finding $a_n$

Finding $b_n$

Find and plot the Fourier series coefficients (magnitude and associated phase) for the following signal with period N. x[n] = 1 + sin((2π/N)n) + 3cos((2π/N)n) + cos((4π/N)n + π/2)

Fourier series coefficients (one period, $k = -2$ to $2$ )

Plots (one period, $k = -2 \dots 2$ , repeating every $N$ )

State and prove the time-scaling property for continuous-time (exponential) Fourier series pair.

Consider a signal x[n] = sin(ω₀n) with period N = 6. Find the Fourier series coefficient of the given signal x[n]. Plot spectra. Also find the average power.

State and prove the time-scaling property for discrete-time Fourier series pair.

Find the frequency domain representation (both magnitude and phase plot) of a periodic signal defined as x[n] = {1, −0.5, 0, 0, 0.5} (x[0] = 1), with period N = 5.

Find the Fourier series coefficients of the signal x[n] = 1 + sin[(2π/N)n] + 3cos[(2π/N)n] + cos[(4π/N)n]

State and prove the time-shifting property for discrete-time Fourier series pair.

State and prove time shifting and time scaling properties of continuous time Fourier series.

Describe the time shifting and frequency shifting properties of continuous time Fourier series. Find Fourier series coefficient of signal x(t) = A cos(ω₀t).

Fourier series coefficients of $x(t) = A\cos(\omega_0 t)$

State and prove frequency shifting property of discrete time Fourier series.

What is Fourier series representation of a signal? Derive an expression for analysis and synthesis equation of the Fourier Series for a CT periodic signal x(t).

What are the significance of Fourier series coefficients of a signal? State and prove time scaling and conjugation properties of continuous time Fourier series representation.

Calculate and plot magnitude and phase spectrum of the following periodic signal. [Figure: sawtooth wave of period 1: x(t) rises linearly from 0 at t = 0 to 1 at t = 1 and then drops back to 0, i.e. x(t) = t for 0 ≤ t < 1, repeating every 1 s; shown from t = −3 to t = 4]

DC term ( $k = 0$ )

Coefficients for $k \ne 0$

Spectra (lines at $\omega = 2\pi k$ )

"Even if a signal is periodic, its Fourier series might not exist." Justify the statement using three different examples.

Derive the formula for synthesis equation and analysis equation of Fourier series. Mention the three conditions that guarantee the existence of Fourier series.

State and explain four properties of discrete time Fourier series.

Write down the condition of convergence of continuous time Fourier series. Find the Fourier series coefficient a_k for the continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2.

Derive the expression of Fourier Series Coefficient in exponential form for Continuous time periodic signal. Describe the condition for existence of Fourier series with suitable examples.

State and prove convolution property of discrete time periodic signals.

A continuous time Fourier Series is evaluated over infinite duration while a discrete time Fourier series is evaluated over one time period only. Why?

Obtain fourier series coefficients of the signal x[n] = 1 + 3cos(2πn/N + π/3), where N is the fundamental time period of x[n]. Also, plot magnitude and phase spectrum of the coefficients.

Derive the expression to compute Fourier series coefficients of exponential Fourier series. Explain Gibbs phenomenon.

DC term ( $k = 0$ )

For $k \ne 0$

State Parseval's relation for discrete time Fourier series. Find the average power of discrete time periodic signal x[n] = sin(πn/3) using Fourier series coefficient.

Average power of $x[n] = \sin(\pi n/3)$

Show that fourier series coefficient of discrete time signal is periodic in nature.

To show: $a_{k+N} = a_k$

Find Fourier series coefficients and associated phase of the signal with fundamental frequency ω₀: x(t) = 1 + sin(ω₀t) + 2cos(ω₀t) + cos(2ω₀t + π/4)

Explain the Dirichlet conditions for convergence of Fourier series.

Find Fourier series coefficient of discrete time periodic signal x[n] = sin ω₀n where ω₀ = 2π/N. Plot the signal when N = 5.

Derive the expression of continuous time exponential Fourier series x[n] = A sin ω₀n

Find Fourier series coefficients of signal and plot the coefficients. [The signal is not printed on the paper.]

State and prove the frequency shifting and convolution properties for discrete time Fourier series pair.

Find the Fourier Series coefficient of the following continuous time periodic signal x(t) = 1.5 for 0 ≤ t < 1; −1.5 for 1 ≤ t < 2, with fundamental frequency π.

State and prove the convolution property of continuous-time Fourier series.

State and prove the linearity and time shift properties of discrete time Fourier series representation.

Find Fourier series coefficients for periodic rectangular pulses with unity amplitude. Draw the magnitude spectrum.

Discuss the following properties of continuous time Fourier series. (a) Time shifting (b) Time scaling (c) Conjugation

How could you represent a signal x(t) with harmonically related exponentials? State and prove conjugation and conjugate symmetry property of CTFS.