Signals and Systems — Chapter 5: Continuous-Time Systems

Asked 8 times
2082 Chaitra · 6 marks
2081 Chaitra · 5 marks
2077 Chaitra · 4 marks
2075 Bhadra · 6 marks
2074 Bhadra · 6 marks
2073 Bhadra · 6 marks
2071 Bhadra · 8 marks
2082 Bhadra (new course) · 4 marks

Derive the convolution integral for continuous time LTI system.

Answer

The convolution integral gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$ in terms of its impulse response $h(t)$ :

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = x(t) * h(t)

Step 1: Represent the input with narrow pulses

Define a unit-area pulse of width $\Delta$ :

\delta_\Delta(t) = \begin{cases} 1/\Delta, & 0 \le t < \Delta \\ 0, & \text{otherwise}\end{cases}

The staircase approximation of $x(t)$ is a sum of shifted, scaled pulses:

\hat{x}(t) = \sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_\Delta(t - k\Delta)\,\Delta

x(t)     __
      __|  |__
   __|        |__     staircase of width-Delta
  |              |    pulses, height x(k.Delta)
--+--+--+--+--+--+--> t
     kD (k+1)D

Step 2: Response to one pulse

Let $\hat{h}_k(t)$ be the response to $\delta_\Delta(t - k\Delta)$ . By linearity (superposition and scaling),

\hat{y}(t) = \sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_k(t)\,\Delta

By time invariance, the response to a shifted pulse is the shifted response: $\hat{h}_k(t) = \hat{h}_0(t - k\Delta)$ .

Step 3: Take the limit $\Delta \to 0$

As $\Delta \to 0$ :

$\hat{x}(t) \to x(t)$ ,
$\delta_\Delta(t) \to \delta(t)$ , so $\hat{h}_0(t) \to h(t)$ , the impulse response,
$k\Delta \to \tau$ (continuous variable), $\Delta \to d\tau$ , and the sum becomes an integral.

\begin{aligned} y(t) &= \lim_{\Delta\to0}\sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_0(t-k\Delta)\,\Delta \\ &= \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau \end{aligned}

The same result follows directly from the sifting property: $x(t) = \int x(\tau)\delta(t-\tau)d\tau$ ; the system maps $\delta(t-\tau) \to h(t-\tau)$ (time invariance) and the integral passes through (linearity).

Meaning and evaluation

The output is a weighted sum of shifted impulse responses; $x(\tau)$ is the weight of the impulse at $\tau$ .
Steps to evaluate: (1) change variable to $\tau$ ; (2) fold $h(\tau)$ to get $h(-\tau)$ ; (3) shift by $t$ to get $h(t-\tau)$ ; (4) multiply by $x(\tau)$ ; (5) integrate over the overlap; repeat for each range of $t$ .
Using $\tau \to t - \tau$ , $y(t) = \int h(\tau)x(t-\tau)d\tau$ also holds (commutative property).

Example: $x(t) = u(t)$ , $h(t) = e^{-at}u(t)$ , $a>0$ : $y(t) = \int_0^t e^{-a(t-\tau)}d\tau = \frac{1}{a}(1 - e^{-at})u(t)$ .

Asked 4 times
2080 Chaitra · 5 marks
2079 Chaitra · 5 marks
2071 Bhadra · 5 marks
2070 Magh · 5 marks

Derive the expression for impulse response of ideal low pass filter (with cutoff frequency ω_c) and discuss.

Answer

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and blocks all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

(zero phase taken for simplicity).

Derivation

Taking the inverse Fourier transform:

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\left[\frac{e^{j\omega t}}{jt}\right]_{-\omega_c}^{\omega_c} = \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} \\ &= \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} = \frac{\sin\omega_c t}{\pi t} \end{aligned}

In sinc form:

h(t) = \frac{\omega_c}{\pi}\cdot\frac{\sin\omega_c t}{\omega_c t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right)

 H(jw)                    h(t)
   1 ______                   wc/pi
    |      |                   /\
    |      |          _  _    /  \    _  _
----+------+--> w   -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc
                       zeros at t = k.pi/wc

Discussion

Peak: at $t = 0$ , $h(0) = \omega_c/\pi$ (limit of $\sin\omega_c t/\pi t$ ).
Zero crossings: at $t = k\pi/\omega_c$ , $k = \pm1, \pm2, \dots$ . A larger $\omega_c$ gives a narrower main lobe (time–bandwidth inverse relation).
Non-causal: $h(t) \ne 0$ for $t < 0$ ; the sinc extends to $-\infty$ . So the filter would have to respond before the impulse is applied, and it is not physically realizable.
Infinite duration and oscillating tails (ringing). Its step response shows overshoot (Gibbs phenomenon, about 9%).
With linear phase $H(j\omega) = e^{-j\omega t_0}$ in the passband, $h(t) = \dfrac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$ : a delayed sinc that is still non-zero for $t<0$ .
Practical filters (Butterworth, Chebyshev, RC) approximate it with a finite transition band.

Asked 3 times
2080 Asoj · 5 marks
2079 Asoj · 5 marks
2078 Chaitra · 5 marks

Evaluate the step response of RC filter.

Answer

The step response $s(t)$ of the RC low-pass filter is its output when $x(t) = u(t)$ and the capacitor is initially uncharged.

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Impulse response

By KVL, with $i(t) = C\,\dfrac{dy}{dt}$ :

x(t) = RC\frac{dy(t)}{dt} + y(t)

Taking the Fourier transform gives $H(j\omega) = \dfrac{1}{1 + j\omega RC}$ , whose inverse transform (using $e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$ ) is

h(t) = \frac{1}{RC}e^{-t/RC}u(t)

Step response

The step response is the running integral of the impulse response:

\begin{aligned} s(t) &= u(t) * h(t) = \int_{-\infty}^{t} h(\tau)\,d\tau \\ &= \int_{0}^{t} \frac{1}{RC}e^{-\tau/RC}\,d\tau \quad (t > 0) \\ &= \frac{1}{RC}\left[-RC\,e^{-\tau/RC}\right]_0^t \\ &= \left(1 - e^{-t/RC}\right)u(t) \end{aligned}

(Check: solving $RC\,\dot y + y = 1$ with $y(0) = 0$ gives the same.)

s(t)
 1 |- - - - - - - - - - - - -
   |           _______-----
.632|- - - -.-'
   |      .'
   |    /
   |  /
   +-/-----+---------------> t
   0      RC

Discussion

$\tau = RC$ is the time constant: $s(RC) = 1 - e^{-1} = 0.632$ .
$s(t)$ reaches about 98% at $4RC$ and 99.3% at $5RC$ .
Rise time (10% to 90%) $= RC\ln 9 \approx 2.2RC$ . Since bandwidth $\omega_c = 1/RC$ , rise time $\approx 2.2/\omega_c$ : a wider bandwidth gives a faster rise.
There is no overshoot (unlike the ideal LPF), and the response is causal because $h(t) = 0$ for $t < 0$ .

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

Describe the frequency response (transfer function) of a continuous time LTI system.

Answer

The frequency response $H(j\omega)$ of a continuous-time LTI system is the Fourier transform of its impulse response. It tells how the system changes the amplitude and phase of each sinusoidal (complex exponential) component of the input.

H(j\omega) = \int_{-\infty}^{\infty} h(t)e^{-j\omega t}\,dt

Eigenfunction property

If $x(t) = e^{j\omega t}$ , the output is

\begin{aligned} y(t) &= \int_{-\infty}^{\infty} h(\tau)e^{j\omega(t-\tau)}\,d\tau \\ &= e^{j\omega t}\int_{-\infty}^{\infty} h(\tau)e^{-j\omega\tau}\,d\tau = H(j\omega)\,e^{j\omega t} \end{aligned}

Complex exponentials pass through unchanged in form, only multiplied by the complex number $H(j\omega)$ .

Transfer function

From the convolution property, $y(t) = x(t)*h(t) \Rightarrow Y(j\omega) = X(j\omega)H(j\omega)$ , so

H(j\omega) = \frac{Y(j\omega)}{X(j\omega)}

For a system described by $\sum_k a_k \frac{d^k y}{dt^k} = \sum_k b_k \frac{d^k x}{dt^k}$ :

H(j\omega) = \frac{\sum_k b_k (j\omega)^k}{\sum_k a_k (j\omega)^k}

Magnitude and phase

$H(j\omega) = |H(j\omega)|\,e^{j\angle H(j\omega)}$

$|H(j\omega)|$ : gain (magnitude response), often plotted in dB on a Bode plot.
$\angle H(j\omega)$ : phase shift; $-d\angle H/d\omega$ is the group delay.
For real $h(t)$ : $|H|$ is even and $\angle H$ is odd in $\omega$ .
A sinusoid $\cos\omega_0 t$ gives output $|H(j\omega_0)|\cos(\omega_0 t + \angle H(j\omega_0))$ .

Example: RC low-pass filter, $H(j\omega) = \dfrac{1}{1 + j\omega RC}$ : $|H| = 1/\sqrt{1 + (\omega RC)^2}$ , $\angle H = -\tan^{-1}\omega RC$ . Low frequencies pass, high frequencies are attenuated.

Asked 2 times
2082 Chaitra · 4 marks
2075 Baisakh · 2+4 marks

Define linear time invariant system. Explain causality and time variance properties of continuous time system.

Answer

LTI system

A linear time-invariant (LTI) system is one that is both

linear: obeys superposition, $a x_1(t) + b x_2(t) \to a y_1(t) + b y_2(t)$ , and
time-invariant: a delay in the input gives the same delay in the output, $x(t - t_0) \to y(t - t_0)$ .

An LTI system is completely described by its impulse response $h(t)$ , and $y(t) = x(t) * h(t)$ .

Causality

A system is causal if the output at any time $t_0$ depends only on present and past inputs ( $t \le t_0$ ), never on future inputs. All real-time physical systems are causal.

Causal: $y(t) = x(t-1)$ , $y(t) = \int_{-\infty}^{t} x(\tau)d\tau$ .
Non-causal: $y(t) = x(t+1)$ , $y(t) = x(-t)$ (at $t = -1$ it needs $x(1)$ ).
For an LTI system: causal $\iff h(t) = 0$ for $t < 0$ .

Time variance

A system is time-invariant if its behaviour does not change with time: if $x(t) \to y(t)$ then $x(t - t_0) \to y(t - t_0)$ for every $t_0$ . Otherwise it is time-variant.

Test: find $y_1(t)$ , the output for $x(t - t_0)$ , and compare with $y(t - t_0)$ .

Example 1: $y(t) = x(2t)$ .

\begin{aligned} y_1(t) &= x(2t - t_0) \\ y(t - t_0) &= x(2(t - t_0)) = x(2t - 2t_0) \end{aligned}

They differ, so the system is time-variant.

Example 2: $y(t) = x(t)^2$ : $y_1(t) = x(t - t_0)^2 = y(t - t_0)$ , so it is time-invariant.

Property	Condition	Example
Causal	output uses $t \le t_0$ only	$y = x(t-2)$
Non-causal	output uses future input	$y = x(t+2)$
Time-invariant	$x(t-t_0) \to y(t-t_0)$	$y = 3x(t)$
Time-variant	shift not preserved	$y = t\,x(t)$

Asked 2 times
2082 Kartik · 5 marks
2083 Baisakh (new course) · 4 marks

Derive the transfer function of a RC low pass filter and plot the magnitude and phase.

Answer

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Transfer function

By KVL, with $i(t) = C\,\dfrac{dy}{dt}$ :

x(t) = RC\frac{dy(t)}{dt} + y(t)

Taking the Fourier transform (with $\frac{d}{dt} \leftrightarrow j\omega$ ):

\begin{aligned} X(j\omega) &= j\omega RC\,Y(j\omega) + Y(j\omega) \\ H(j\omega) &= \frac{Y(j\omega)}{X(j\omega)} = \frac{1}{1 + j\omega RC} \end{aligned}

In Laplace form, $H(s) = \dfrac{1}{1 + sRC}$ . (Same result from the voltage divider: $H = \dfrac{1/j\omega C}{R + 1/j\omega C}$ .)

Magnitude and phase

|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}, \qquad \angle H(j\omega) = -\tan^{-1}(\omega RC)

$\omega$	$\lvert H\rvert$	$\lvert H\rvert$ (dB)	$\angle H$
$0$	$1$	$0$	$0^\circ$
$1/RC$	$1/\sqrt2 = 0.707$	$-3$	$-45^\circ$
$10/RC$	$0.0995$	$-20$	$-84.3^\circ$
$\to\infty$	$\to 0$	slope $-20$ dB/dec	$\to -90^\circ$

|H|                         angle H
 1 |----.                  0 |----.
   |     `.                   |     `.
.707|- - - -*                -45|- - - -*
   |         `-.              |         `-.___
   |             `---.    -90 |- - - - - - - -
   +------+-----------> w     +------+--------> w
        1/RC                       1/RC

The cutoff (half-power, $-3$ dB) frequency is $\omega_c = 1/RC$ ; the circuit is a first-order low-pass filter.

Asked 2 times
2082 Kartik · 5 marks
2071 Magh · 8 marks

Find the output y(t) of LTI system if the impulse response h(t) = e^(−t)u(t−3) and input x(t) = u(t−2).

Answer

For an LTI system, $y(t) = x(t) * h(t) = \displaystyle\int_{-\infty}^{\infty} x(\tau)\,h(t - \tau)\,d\tau$ .

Given $x(t) = u(t-2)$ and $h(t) = e^{-t}u(t-3)$ .

Step 1: Write both functions in $\tau$

x(\tau) = u(\tau - 2), \qquad h(t-\tau) = e^{-(t-\tau)}\,u(t - \tau - 3)

$x(\tau) \neq 0$ when $\tau > 2$ .
$h(t-\tau) \neq 0$ when $t - \tau > 3$ , i.e. $\tau < t - 3$ .

 x(tau)=1 for tau>2      h(t-tau) for tau<t-3
          ___________   ___________
         |              decaying   |
 --------+-------->   <------------+------ tau
         2                        t-3
 overlap: 2 < tau < t-3  (exists only if t-3 > 2)

Step 2: Range of $t$

The overlap $2 < \tau < t - 3$ exists only when $t - 3 > 2$ , i.e. $t > 5$ .

For $t < 5$ : no overlap, $y(t) = 0$ .

Step 3: Evaluate for $t > 5$

\begin{aligned} y(t) &= \int_{2}^{t-3} 1\cdot e^{-(t-\tau)}\,d\tau = e^{-t}\int_{2}^{t-3} e^{\tau}\,d\tau \\ &= e^{-t}\left[e^{\tau}\right]_{2}^{t-3} = e^{-t}\left(e^{t-3} - e^{2}\right) \\ &= e^{-3} - e^{-(t-2)} \end{aligned}

Result

y(t) = \left(e^{-3} - e^{-(t-2)}\right)u(t-5)

Answer: $y(t) = 0$ for $t < 5$ , and $y(t) = e^{-3} - e^{-(t-2)}$ for $t \ge 5$ .

Checks:

At $t = 5$ : $e^{-3} - e^{-3} = 0$ (continuous start).
As $t \to \infty$ : $y \to e^{-3} \approx 0.0498$ .
Shift view: $u(t)*e^{-t}u(t) = (1 - e^{-t})u(t)$ ; here $h(t) = e^{-3}\,e^{-(t-3)}u(t-3)$ , so total shift is $2 + 3 = 5$ : $y(t) = e^{-3}(1 - e^{-(t-5)})u(t-5)$ , which is the same expression.

Asked 2 times
2081 Chaitra · 6 marks
2073 Magh · 4+2 marks

Derive the expression for impulse response of RC filter and plot the magnitude and phase response.

Answer

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Impulse response

By KVL, with $i(t) = C\,\dfrac{dy}{dt}$ :

x(t) = RC\frac{dy(t)}{dt} + y(t)

Take the Fourier transform:

\begin{aligned} X(j\omega) &= j\omega RC\,Y(j\omega) + Y(j\omega) \\ H(j\omega) &= \frac{Y(j\omega)}{X(j\omega)} = \frac{1}{1 + j\omega RC} \end{aligned}

Write it as $H(j\omega) = \dfrac{1/RC}{1/RC + j\omega}$ and use the pair $e^{-at}u(t) \leftrightarrow \dfrac{1}{a + j\omega}$ with $a = 1/RC$ :

h(t) = \frac{1}{RC}\,e^{-t/RC}\,u(t)

(Physically: an impulse charges the capacitor instantly to $1/RC$ , which then discharges through $R$ with time constant $RC$ .)

h(t)
1/RC|.
    | `.
    |   `-.
    |      `--.___
    +-------+-------`----> t
    0      RC

$h(t) = 0$ for $t < 0$ , so the filter is causal; $\int |h|\,dt = 1$ , so it is stable.

Magnitude and phase response

|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}, \qquad \angle H(j\omega) = -\tan^{-1}(\omega RC)

$\omega$	$\lvert H\rvert$	$\lvert H\rvert$ (dB)	$\angle H$
$0$	$1$	$0$	$0^\circ$
$1/RC$	$1/\sqrt2 = 0.707$	$-3$	$-45^\circ$
$10/RC$	$0.0995$	$-20$	$-84.3^\circ$
$\to\infty$	$\to 0$	slope $-20$ dB/dec	$\to -90^\circ$

|H|                         angle H
 1 |----.                  0 |----.
   |     `.                   |     `.
.707|- - - -*                -45|- - - -*
   |         `-.              |         `-.___
   |             `---.    -90 |- - - - - - - -
   +------+-----------> w     +------+--------> w
        1/RC                       1/RC

The cutoff (half-power, $-3$ dB) frequency is $\omega_c = 1/RC$ ; the circuit is a first-order low-pass filter.

Asked 2 times
2080 Asoj · 6 marks
2079 Asoj · 8 marks

Find the convolution between the signals x(t) = 1 for −1 < t < 1, 0 otherwise and h(t) = 1 for −1 < t < 1, 0 otherwise.

Answer

$y(t) = x(t) * h(t) = \displaystyle\int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$ , with $x(\tau) = 1$ for $-1 < \tau < 1$ , and $h(t - \tau) = 1$ for $-1 < t - \tau < 1$ , i.e. $t - 1 < \tau < t + 1$ .

The result is non-zero only where the two pulses overlap. Each has width 2, so the output extends from $t = -2$ to $t = 2$ .

       x(tau)            h(t-tau)
     _________        _________
    |         |      |         |
 ---+----+----+--  --+----+----+--> tau
   -1    0    1    t-1    t   t+1

Case 1: $t < -2$

$t + 1 < -1$ : no overlap, $y(t) = 0$ .

Case 2: $-2 \le t < 0$ (partial overlap from left)

Overlap from $\tau = -1$ to $\tau = t + 1$ :

y(t) = \int_{-1}^{t+1} 1\,d\tau = t + 2

Case 3: $0 \le t < 2$ (partial overlap from right)

Overlap from $\tau = t - 1$ to $\tau = 1$ :

y(t) = \int_{t-1}^{1} 1\,d\tau = 2 - t

Case 4: $t \ge 2$

$t - 1 > 1$ : no overlap, $y(t) = 0$ .

Result

y(t) = \begin{cases} t + 2, & -2 \le t < 0 \\ 2 - t, & 0 \le t < 2 \\ 0, & |t| \ge 2 \end{cases} \;=\; (2 - |t|)\ \text{for } |t| < 2

 y(t)
   2 |      /\
     |     /  \
   1 |    /    \
     |   /      \
  ---+--/---+----\---> t
      -2    0     2

Answer: a triangular pulse of height 2 at $t = 0$ , base from $-2$ to $2$ . Convolving two equal rectangles always gives a triangle whose width is the sum of the widths (2 + 2 = 4) and whose peak equals the area of overlap (2).

Asked 2 times
2079 Chaitra · 2+5 marks
2070 Bhadra · 2+3 marks

What are the properties of system? Determine whether the given system is time-variant or not. y(t) = sin[x(t)]

Answer

Properties of systems

Memory: memoryless if $y(t)$ depends only on $x(t)$ at the same instant (e.g. $y = 2x(t)$ ); otherwise has memory (e.g. $y = x(t-1)$ ).
Causality: causal if $y(t)$ depends only on present and past inputs.
Linearity: satisfies superposition: $a x_1 + b x_2 \to a y_1 + b y_2$ .
Time invariance: $x(t - t_0) \to y(t - t_0)$ .
Stability (BIBO): every bounded input gives a bounded output.
Invertibility: distinct inputs give distinct outputs, so an inverse system exists.

Is $y(t) = \sin[x(t)]$ time-variant?

Let the input be delayed: $x_1(t) = x(t - t_0)$ . The output for this input is

y_1(t) = \sin[x_1(t)] = \sin[x(t - t_0)]

The original output delayed by $t_0$ is

y(t - t_0) = \sin[x(t - t_0)]

Since $y_1(t) = y(t - t_0)$ , the system is time-invariant (not time-variant). The operation $\sin[\cdot]$ does not depend on $t$ explicitly.

Other properties, for completeness:

Property	Result	Reason
Time invariance	Time-invariant	shown above
Linearity	Non-linear	$\sin(x_1 + x_2) \ne \sin x_1 + \sin x_2$
Memory	Memoryless	uses only $x(t)$
Causality	Causal	no future input
Stability	Stable	$\lvert y(t)\rvert \le 1$ for any input
Invertibility	Not invertible	$x$ and $x + 2\pi$ give same $y$

Asked 2 times
2079 Chaitra · 4 marks
2071 Magh · 4 marks

Describe bode plot with example.

Answer

A Bode plot is a pair of graphs of a system's frequency response $H(j\omega)$ against $\log_{10}\omega$ :

Magnitude plot: $20\log_{10}|H(j\omega)|$ in dB,
Phase plot: $\angle H(j\omega)$ in degrees.

Because a product of factors becomes a sum of dB values, the plot is drawn by adding simple straight-line (asymptotic) approximations of each factor.

Basic factors

Factor	Magnitude (asymptote)	Phase
Constant $K$	$20\log K$ flat	$0^\circ$
Zero at origin $j\omega$	$+20$ dB/dec, 0 dB at $\omega=1$	$+90^\circ$
Pole at origin $1/j\omega$	$-20$ dB/dec	$-90^\circ$
Simple pole $\frac{1}{1 + j\omega/a}$	0 dB up to $a$ , then $-20$ dB/dec	$0^\circ \to -90^\circ$ , $-45^\circ$ at $a$
Simple zero $1 + j\omega/a$	0 dB up to $a$ , then $+20$ dB/dec	$0^\circ \to +90^\circ$

The frequency $a$ is the corner (break) frequency; the true curve is 3 dB from the asymptote there.

Example

H(j\omega) = \frac{10}{1 + j\omega/10}

$K = 10 \Rightarrow 20\log 10 = 20$ dB.
Pole at $\omega = 10$ rad/s.

Magnitude: 20 dB flat up to 10 rad/s, then falls at $-20$ dB/dec: 0 dB at 100 rad/s, $-20$ dB at 1000 rad/s. Actual value at $\omega = 10$ : $20 - 3 = 17$ dB.

Phase: $-\tan^{-1}(\omega/10)$ : about $0^\circ$ at 1, $-45^\circ$ at 10, about $-90^\circ$ at 100 rad/s.

dB
 20 |--------.
    |         \   -20 dB/dec
  0 +----------\--------------> w (log)
    1     10   100   1000
deg
  0 |----.
-45 |     `--.
-90 |         `-------------> w (log)
    1     10   100

Uses

Shows bandwidth, cutoff frequency and filter type at a glance.
Gives gain margin and phase margin for stability of feedback systems.
Quick to sketch by hand, covering a wide frequency range.

2082 Chaitra · 2+4 marks

Define frequency response of the discrete LTI system. Derive formula to calculate the impulse response of continuous time ideal low pass filter.

Answer

Frequency response of a discrete-time LTI system

The frequency response $H(e^{j\Omega})$ of a discrete-time LTI system is the DTFT of its impulse response $h[n]$ :

H(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} h[n]\,e^{-j\Omega n}

If $x[n] = e^{j\Omega n}$ , then $y[n] = H(e^{j\Omega})\,e^{j\Omega n}$ , so $H(e^{j\Omega})$ gives the gain $|H|$ and phase shift $\angle H$ at each frequency. Also $Y(e^{j\Omega}) = H(e^{j\Omega})X(e^{j\Omega})$ . It is always periodic in $\Omega$ with period $2\pi$ .

Example: $h[n] = a^n u[n]$ , $|a| < 1$ gives $H(e^{j\Omega}) = \dfrac{1}{1 - a e^{-j\Omega}}$ .

Impulse response of the CT ideal low-pass filter

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

The impulse response is the inverse Fourier transform of $H(j\omega)$ :

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned}

 H(jw)                     h(t)
   1 ______                    wc/pi
    |      |                    /\
    |      |           _  _    /  \    _  _
----+------+--> w    -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc

Peak value $h(0) = \omega_c/\pi$ ; zeros at $t = k\pi/\omega_c$ , $k \ne 0$ .
$h(t) \ne 0$ for $t < 0$ , so the ideal LPF is non-causal and cannot be built exactly.

2079 Jestha · 2+8 marks

What is an LTI system? Find the output of the LTI system with impulse response h(t) = e^(−2t)u(t) when the input to the system is x(t) = 1 for |t| < 5; 0 otherwise.

Answer

LTI system

An LTI system is a system that is both linear (obeys superposition) and time-invariant (a shift in input gives the same shift in output). It is fully described by its impulse response $h(t)$ , and the output for any input is the convolution $y(t) = x(t) * h(t)$ .

Output for $h(t) = e^{-2t}u(t)$ , $x(t) = 1$ for $|t| < 5$

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

$x(\tau) = 1$ for $-5 < \tau < 5$ .
$h(t - \tau) = e^{-2(t-\tau)}$ for $t - \tau > 0$ , i.e. $\tau < t$ .

So the integrand is non-zero for $-5 < \tau < \min(t, 5)$ .

   x(tau)                  h(t-tau)
    ___________          .-'|
   |           |     .-'    |
 --+-----+-----+--.--------+---> tau
  -5     0     5           t

Case 1: $t < -5$ . No overlap.

y(t) = 0

Case 2: $-5 \le t < 5$ . Overlap from $-5$ to $t$ .

\begin{aligned} y(t) &= \int_{-5}^{t} e^{-2(t-\tau)}\,d\tau = e^{-2t}\left[\frac{e^{2\tau}}{2}\right]_{-5}^{t} \\ &= \frac{e^{-2t}}{2}\left(e^{2t} - e^{-10}\right) = \frac{1}{2}\left(1 - e^{-2(t+5)}\right) \end{aligned}

Case 3: $t \ge 5$ . Overlap from $-5$ to $5$ .

\begin{aligned} y(t) &= \int_{-5}^{5} e^{-2(t-\tau)}\,d\tau = \frac{e^{-2t}}{2}\left(e^{10} - e^{-10}\right) \\ &= \frac{1}{2}\left(e^{-2(t-5)} - e^{-2(t+5)}\right) \end{aligned}

Result

y(t) = \begin{cases} 0, & t < -5 \\[2pt] \dfrac12\left(1 - e^{-2(t+5)}\right), & -5 \le t < 5 \\[6pt] \dfrac12\left(e^{-2(t-5)} - e^{-2(t+5)}\right), & t \ge 5 \end{cases}

 y(t)
 0.5 |      .------------.
     |    /               \
     |   /                  `.
  ---+--/--------+-----------+-`---___--> t
       -5        0           5

Checks: $y(-5) = 0$ ; at $t = 5$ both pieces give $\tfrac12(1 - e^{-20}) \approx 0.5$ (continuous); $y \to 0$ as $t \to \infty$ . Physically, the output rises toward 0.5 with time constant 0.5 s while the input is on, then decays after $t = 5$ .

2079 Jestha · 7 marks

Show that ideal low pass filter is not practically realizable.

Answer

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

A physically realizable system must be causal: for an LTI system this means $h(t) = 0$ for $t < 0$ .

Step 1: Impulse response

The impulse response is the inverse Fourier transform of $H(j\omega)$ :

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned}

 H(jw)                     h(t)
   1 ______                    wc/pi
    |      |                    /\
    |      |           _  _    /  \    _  _
----+------+--> w    -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc

Step 2: Check causality

$h(t)$ is an even sinc function. It has its peak $\omega_c/\pi$ at $t = 0$ and oscillating tails on both sides, extending to $t = -\infty$ . For example, at $t = -\pi/(2\omega_c)$ :

h\!\left(-\frac{\pi}{2\omega_c}\right) = \frac{\sin(-\pi/2)}{\pi(-\pi/2\omega_c)} = \frac{2\omega_c}{\pi^2} \neq 0

So $h(t) \ne 0$ for $t < 0$ : the filter must produce output before the impulse is applied at $t = 0$ .

Step 3: Delay does not help

With linear phase, $H(j\omega) = e^{-j\omega t_0}$ for $|\omega| < \omega_c$ , giving

h(t) = \frac{\sin\omega_c(t - t_0)}{\pi(t - t_0)}

This is centred at $t_0$ , but its tail still extends to $-\infty$ , so $h(t) \ne 0$ for some $t < 0$ for every finite $t_0$ .

Step 4: Paley–Wiener criterion

A causal filter with magnitude $|H(j\omega)|$ must satisfy

\int_{-\infty}^{\infty}\frac{\big|\ln|H(j\omega)|\big|}{1 + \omega^2}\,d\omega < \infty

For the ideal LPF, $|H| = 0$ over the whole stopband, so $|\ln|H|| = \infty$ there and the integral diverges. A causal filter can have zero gain only at isolated frequencies, never over a band.

Hence the ideal low-pass filter is non-causal and not physically realizable. Practical filters such as Butterworth and Chebyshev only approximate it, accepting a finite transition band, some passband ripple or stopband leakage, and a non-linear phase.

2078 Baisakh · 6 marks

What do you mean by impulse response of a LTI system? Derive formula to calculate impulse response of continuous time ideal low pass filter.

Answer

Impulse response of an LTI system

The impulse response $h(t)$ of an LTI system is its output when the input is the unit impulse $\delta(t)$ and the system is initially at rest:

\delta(t) \;\to\; h(t)

Because any input can be written as $x(t) = \int x(\tau)\delta(t-\tau)d\tau$ , linearity and time invariance give $y(t) = \int x(\tau)h(t-\tau)d\tau = x(t)*h(t)$ . So $h(t)$ completely characterises an LTI system: causality ( $h(t) = 0$ for $t<0$ ), stability ( $\int|h|\,dt < \infty$ ) and frequency response ( $H(j\omega) = \mathcal{F}\{h(t)\}$ ) can all be read from it.

Impulse response of the CT ideal low-pass filter

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

The impulse response is the inverse Fourier transform of $H(j\omega)$ :

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned}

 H(jw)                     h(t)
   1 ______                    wc/pi
    |      |                    /\
    |      |           _  _    /  \    _  _
----+------+--> w    -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc

$h(0) = \omega_c/\pi$ ; zero crossings at $t = k\pi/\omega_c$ ( $k = \pm1, \pm2, \dots$ ).
Width of the main lobe $2\pi/\omega_c$ shrinks as bandwidth $\omega_c$ grows.
$h(t)$ exists for $t < 0$ , so the ideal LPF is non-causal and not realizable; it is also of infinite duration.

2078 Baisakh · 3 marks

State and prove commutative property of continuous time LTI system.

Answer

Statement: Convolution of continuous-time signals is commutative:

x(t) * h(t) = h(t) * x(t)

For LTI systems, this means the output is the same whether the input is $x(t)$ applied to a system with impulse response $h(t)$ , or $h(t)$ applied to a system with impulse response $x(t)$ .

Proof:

x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t - \tau)\,d\tau

Substitute $\lambda = t - \tau$ , so $\tau = t - \lambda$ and $d\tau = -d\lambda$ . When $\tau \to -\infty$ , $\lambda \to \infty$ ; when $\tau \to \infty$ , $\lambda \to -\infty$ :

\begin{aligned} x(t) * h(t) &= \int_{\infty}^{-\infty} x(t - \lambda)\,h(\lambda)\,(-d\lambda) \\ &= \int_{-\infty}^{\infty} h(\lambda)\,x(t - \lambda)\,d\lambda = h(t) * x(t) \end{aligned}

$\blacksquare$

x(t) --> [ h(t) ] --> y(t)   ==   h(t) --> [ x(t) ] --> y(t)

Use: we can flip whichever signal is simpler when evaluating the integral; and for cascaded LTI systems, the order of the systems can be swapped.

2078 Baisakh · 5 marks

Explain invertibility of LTI system and derive the necessary condition for two LTI systems to become inverse of each other.

Answer

A system is invertible if distinct inputs produce distinct outputs, so that the input can be recovered from the output. The system that recovers it is the inverse system.

x(t) --> [ h(t) ] --w(t)--> [ h1(t) ] --> y(t) = x(t)

Condition for LTI systems

Let the system have impulse response $h(t)$ and the inverse system $h_1(t)$ . The cascade is an LTI system with impulse response $h(t) * h_1(t)$ . For the cascade to be the identity system ( $y(t) = x(t)$ for every $x$ ):

\begin{aligned} y(t) &= x(t) * \big[h(t) * h_1(t)\big] = x(t) \\ \Rightarrow\; h(t) * h_1(t) &= \delta(t) \end{aligned}

because $x(t) * \delta(t) = x(t)$ is the only identity for convolution.

In the frequency domain (convolution property):

H(j\omega)\,H_1(j\omega) = 1 \;\Rightarrow\; H_1(j\omega) = \frac{1}{H(j\omega)}

So an LTI system is invertible only if $H(j\omega) \ne 0$ at all frequencies (no information destroyed). Similarly, in discrete time: $h[n] * h_1[n] = \delta[n]$ .

Examples

Delay: $h(t) = \delta(t - t_0)$ . Inverse: $h_1(t) = \delta(t + t_0)$ , since $\delta(t - t_0) * \delta(t + t_0) = \delta(t)$ .
Accumulator: $y[n] = \sum_{k=-\infty}^{n} x[k]$ , $h[n] = u[n]$ . Inverse: first difference $y[n] = x[n] - x[n-1]$ , $h_1[n] = \delta[n] - \delta[n-1]$ . Check: $u[n] - u[n-1] = \delta[n]$ .
Not invertible: $y(t) = 0$ , or an ideal LPF ( $H = 0$ in stopband; those frequencies are lost).

2078 Poush · 3 marks

Elaborate the physical meaning of convolution integral using suitable example.

Answer

The convolution integral

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

says that the output of an LTI system at time $t$ is the sum of the responses to all past (and present) input values, each weighted by how strongly the system still "remembers" it.

Physical meaning:

The input is thought of as a chain of very narrow impulses; the impulse at time $\tau$ has strength $x(\tau)\,d\tau$ .
Each impulse produces its own response $h(t-\tau)$ , which starts at $\tau$ and dies out according to the system's memory.
The output is the superposition (linearity) of these shifted copies (time invariance).
$h(t - \tau)$ acts as a weighting function: $h$ of a large age $t - \tau$ is small for a system with short memory.

Example – RC circuit charging: $h(t) = \frac{1}{RC}e^{-t/RC}u(t)$ . The capacitor voltage now is

y(t) = \int_{-\infty}^{t} x(\tau)\frac{1}{RC}e^{-(t-\tau)/RC}\,d\tau

Recent inputs (small $t - \tau$ ) count almost fully; inputs older than about $5RC$ have almost no effect, because that charge has leaked away. Another everyday example is an echo in a hall: what you hear is the present sound plus fading copies of earlier sounds.

2078 Poush · 6+6 marks

Derive convolution sum. Prove that an ideal low-pass filter is a non-causal system.

Answer

Derivation of the convolution sum

Any discrete-time signal can be written as a sum of shifted, scaled unit impulses (sifting property):

x[n] = \sum_{k=-\infty}^{\infty} x[k]\,\delta[n-k]

e.g. $x[n] = \{2, 3\}$ (starting at $n=0$ ) is $2\delta[n] + 3\delta[n-1]$ .

Let the system be LTI with impulse response $h[n]$ , i.e. $\delta[n] \to h[n]$ .

Time invariance: $\delta[n-k] \to h[n-k]$ .
Scaling (homogeneity): $x[k]\,\delta[n-k] \to x[k]\,h[n-k]$ ( $x[k]$ is just a number for fixed $k$ ).
Additivity: sum over all $k$ :

y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] = x[n] * h[n]

This is the convolution sum. By substituting $m = n - k$ , $y[n] = \sum_m h[m]x[n-m]$ (commutative).

Procedure: fold $h[k]$ to $h[-k]$ , shift by $n$ , multiply with $x[k]$ , add the products; repeat for each $n$ .

Example: $x[n] = \{1, 2\}$ , $h[n] = \{1, 1\}$ (both starting at 0): $y[0] = 1$ , $y[1] = 1\cdot1 + 2\cdot1 = 3$ , $y[2] = 2$ , so $y[n] = \{1, 3, 2\}$ .

Ideal LPF is non-causal

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

Impulse response: The impulse response is the inverse Fourier transform of $H(j\omega)$ :

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned}

 H(jw)                     h(t)
   1 ______                    wc/pi
    |      |                    /\
    |      |           _  _    /  \    _  _
----+------+--> w    -' '-' '-/----\-' '-' '-> t
  -wc  0   wc               -pi/wc pi/wc

Causality test: an LTI system is causal if and only if $h(t) = 0$ for $t < 0$ (the output cannot depend on future input, since $y(t) = \int h(\tau)x(t - \tau)d\tau$ uses $x$ at $t - \tau > t$ whenever $h(\tau) \ne 0$ for $\tau < 0$ ).

For the ideal LPF, $h(t) = \dfrac{\sin\omega_c t}{\pi t}$ is an even function, non-zero on both sides of $t = 0$ . For example,

h\!\left(-\frac{\pi}{2\omega_c}\right) = \frac{2\omega_c}{\pi^2} \ne 0

So $h(t) \ne 0$ for $t < 0$ and the ideal LPF is non-causal. Even with a delay $t_0$ , $h(t) = \dfrac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$ still has a tail reaching $t = -\infty$ , so no finite delay makes it causal. (The discrete-time ideal LPF, $h[n] = \dfrac{\sin\Omega_c n}{\pi n}$ , is non-causal for the same reason.) Hence it cannot be realized in real time.

2077 Chaitra · 2+3 marks

What do you mean by causal and non causal systems? Derive the condition for a continuous time LTI system to be causal.

Answer

Causal and non-causal systems

A causal system is one whose output at any time depends only on the present and past values of the input, not on future values. Example: $y(t) = x(t) + 2x(t-1)$ . All physical real-time systems are causal.
A non-causal system's output depends on future input values as well. Example: $y(t) = x(t+1)$ , $y(t) = x(-t)$ . Such systems can be used only on stored (recorded) data, e.g. image processing.

Condition for a CT LTI system to be causal

For an LTI system,

y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau

Split the integral at $\tau = 0$ :

y(t) = \underbrace{\int_{0}^{\infty} h(\tau)\,x(t-\tau)\,d\tau}_{\text{uses } x \text{ at times} \le t} + \underbrace{\int_{-\infty}^{0} h(\tau)\,x(t-\tau)\,d\tau}_{\text{uses } x \text{ at times} > t}

The second integral uses future inputs $x(t - \tau)$ with $t - \tau > t$ . For $y(t)$ to be independent of every future input, this term must vanish for all $x$ , which requires

h(t) = 0 \quad \text{for } t < 0

Then

y(t) = \int_{0}^{\infty} h(\tau)\,x(t-\tau)\,d\tau = \int_{-\infty}^{t} x(\tau)\,h(t-\tau)\,d\tau

Condition: a CT LTI system is causal if and only if its impulse response $h(t) = 0$ for $t < 0$ . (Physically, the system cannot respond before the impulse is applied at $t = 0$ .)

Examples: $h(t) = e^{-2t}u(t)$ is causal; $h(t) = e^{-2|t|}$ or $h(t) = \sin(\omega_c t)/\pi t$ is non-causal.

2077 Chaitra · 3+3 marks

Prove, with necessary derivations, that an ideal low pass filter is not physically realizable. Also derive the expression for step response of ideal low pass filter.

Answer

Ideal LPF is not physically realizable

The ideal low-pass filter passes all frequencies below the cutoff $\omega_c$ with unit gain and rejects all others:

H(j\omega) = \begin{cases} 1, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

Its impulse response is the inverse Fourier transform: The impulse response is the inverse Fourier transform of $H(j\omega)$ :

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(j\omega)e^{j\omega t}\,d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c t} - e^{-j\omega_c t}}{jt} = \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c t}{jt} \\ &= \frac{\sin\omega_c t}{\pi t} = \frac{\omega_c}{\pi}\,\text{sinc}\!\left(\frac{\omega_c t}{\pi}\right) \end{aligned}

A realizable (causal) LTI system needs $h(t) = 0$ for $t < 0$ . Here $h(t)$ is an even sinc, non-zero on both sides; e.g. $h(-\pi/2\omega_c) = 2\omega_c/\pi^2 \ne 0$ . It responds before the impulse arrives, so it is non-causal. Adding delay $t_0$ gives $\dfrac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}$ , still non-zero for $t < 0$ . Also, $|H| = 0$ over a whole band violates the Paley–Wiener condition $\int \frac{|\ln|H(j\omega)||}{1+\omega^2}d\omega < \infty$ . Hence it is not physically realizable.

Step response of the ideal LPF

The step response is the running integral of $h(t)$ :

\begin{aligned} s(t) &= \int_{-\infty}^{t} h(\tau)\,d\tau = \int_{-\infty}^{t}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau \\ &= \int_{-\infty}^{0}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau + \int_{0}^{t}\frac{\sin\omega_c\tau}{\pi\tau}\,d\tau \end{aligned}

Using $\int_0^\infty \frac{\sin\omega_c\tau}{\tau}d\tau = \frac{\pi}{2}$ , the first integral is $\frac{1}{\pi}\cdot\frac{\pi}{2} = \frac12$ . In the second, put $x = \omega_c\tau$ :

s(t) = \frac12 + \frac{1}{\pi}\int_0^{\omega_c t}\frac{\sin x}{x}\,dx = \frac12 + \frac{1}{\pi}\,\text{Si}(\omega_c t)

where $\text{Si}(z) = \int_0^z \frac{\sin x}{x}dx$ is the sine integral.

 s(t)        overshoot ~9%
  1  |          _/\_ _
     |         /     ' '-----
 0.5 |- - - - /
     |  _    /
  0  +-' '--/---------------> t
           0   pi/wc

$s(0) = 0.5$ ; $s(-\infty) = 0$ ; $s(\infty) = 1$ .
Maximum at $t = \pi/\omega_c$ : $s = 0.5 + \text{Si}(\pi)/\pi \approx 1.09$ (about 9% overshoot, Gibbs phenomenon).
The response starts before $t = 0$ (non-causal) and rises from 0 to 1 in a time of order $\pi/\omega_c$ .

2077 Chaitra · 5 marks

Consider a continuous time system with input x(t) and output y(t) related by y(t) = x(t−2) + x(2−t). Is the system Linear Time invariant? Explain.

Answer

Given $y(t) = x(t-2) + x(2-t)$ . Check linearity and time invariance separately.

Linearity

Let $x_1(t) \to y_1(t) = x_1(t-2) + x_1(2-t)$ and $x_2(t) \to y_2(t) = x_2(t-2) + x_2(2-t)$ .

For $x_3(t) = a x_1(t) + b x_2(t)$ :

\begin{aligned} y_3(t) &= x_3(t-2) + x_3(2-t) \\ &= a x_1(t-2) + b x_2(t-2) + a x_1(2-t) + b x_2(2-t) \\ &= a\big[x_1(t-2) + x_1(2-t)\big] + b\big[x_2(t-2) + x_2(2-t)\big] \\ &= a y_1(t) + b y_2(t) \end{aligned}

Superposition holds, so the system is linear.

Time invariance

Apply a delayed input $x_1(t) = x(t - t_0)$ . The output is obtained by replacing $x(\cdot)$ with $x(\cdot - t_0)$ :

y_1(t) = x_1(t-2) + x_1(2-t) = x(t - 2 - t_0) + x(2 - t - t_0)

Now delay the original output by $t_0$ (replace $t$ by $t - t_0$ ):

y(t - t_0) = x(t - t_0 - 2) + x(2 - t + t_0)

The first terms match, but the second terms differ: $x(2 - t - t_0) \ne x(2 - t + t_0)$ . So $y_1(t) \ne y(t - t_0)$ and the system is time-variant.

Counter-example: let $x(t) = \delta(t)$ . Then $y(t) = \delta(t-2) + \delta(2-t) = 2\delta(t-2)$ . For $x(t) = \delta(t-1)$ : $y(t) = \delta(t-3) + \delta(1-t) = \delta(t-3) + \delta(t-1)$ , which is not $y(t-1) = 2\delta(t-3)$ .

The cause is the time-reversal term $x(2 - t)$ : a delay of the input moves this part of the output earlier, not later.

Answer: the system is linear but time-variant, so it is not an LTI system. (It is also non-causal: $y(0) = x(-2) + x(2)$ needs a future input.)

2076 Baisakh · 3 marks

Write a short note on LTI system.

Answer

A linear time-invariant (LTI) system is a system that satisfies both:

Linearity (superposition): $a x_1(t) + b x_2(t) \to a y_1(t) + b y_2(t)$ .
Time invariance: $x(t - t_0) \to y(t - t_0)$ .

Key results:

An LTI system is completely described by its impulse response $h(t)$ (or $h[n]$ ). The output for any input is the convolution

y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau, \qquad y[n] = \sum_k x[k]h[n-k]

In frequency domain, $Y(j\omega) = H(j\omega)X(j\omega)$ ; complex exponentials are eigenfunctions.
Causal iff $h(t) = 0$ for $t < 0$ . BIBO stable iff $\int |h(t)|\,dt < \infty$ .
Cascade: $h_1 * h_2$ (order can be swapped). Parallel: $h_1 + h_2$ .
Described by linear constant-coefficient differential/difference equations.

Examples: RC circuit, ideal delay, moving-average filter $y[n] = \frac12(x[n] + x[n-1])$ .

2076 Baisakh · 3 marks

Write a short note on Bode plot.

Answer

A Bode plot is a pair of graphs of a system's frequency response $H(j\omega)$ against $\log_{10}\omega$ :

Magnitude plot: $20\log_{10}|H(j\omega)|$ in dB,
Phase plot: $\angle H(j\omega)$ in degrees.

Because a product of factors becomes a sum of dB values, the plot is drawn by adding simple straight-line (asymptotic) approximations of each factor.

Rules for sketching:

A constant $K$ adds $20\log K$ dB.
Each pole at the origin adds $-20$ dB/decade and $-90^\circ$ ; each zero at the origin $+20$ dB/dec and $+90^\circ$ .
A simple pole $1/(1 + j\omega/a)$ : 0 dB up to the corner frequency $a$ , then $-20$ dB/dec; phase goes from $0^\circ$ to $-90^\circ$ ( $-45^\circ$ at $a$ ). The true curve is 3 dB below the corner.

Example: $H(j\omega) = \dfrac{1}{1 + j\omega RC}$ (RC LPF): flat 0 dB up to $\omega = 1/RC$ , then $-20$ dB/dec.

Uses: read bandwidth and filter type, and find gain and phase margins for stability of control systems.

2076 Bhadra · 5+6 marks

Derive formula to calculate convolution integral for continuous time LTI system. Find convolution between following continuous time signals x(t) = e^(−t)u(t) and y(t) = e^(−2t)u(t−3)

Answer

Derivation of the convolution integral

The convolution integral gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$ , once the impulse response $h(t)$ is known.

Step 1 – Represent the input by impulses (sifting property):

x(t) = \int_{-\infty}^{\infty} x(\tau)\,\delta(t-\tau)\,d\tau

So $x(t)$ is a continuous sum of shifted impulses $\delta(t-\tau)$ , each weighted by $x(\tau)\,d\tau$ .

Step 2 – Response to one impulse: by definition, $\delta(t) \rightarrow h(t)$ .

Step 3 – Time invariance: $\delta(t-\tau) \rightarrow h(t-\tau)$ .

Step 4 – Linearity (homogeneity): $x(\tau)\,\delta(t-\tau)\,d\tau \rightarrow x(\tau)\,h(t-\tau)\,d\tau$ .

Step 5 – Linearity (additivity): adding (integrating) the responses to all the weighted impulses,

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = x(t) * h(t)

This is the convolution integral. By changing the variable $\lambda = t-\tau$ it can also be written $y(t)=\int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$ .

Graphical steps: (1) change $t$ to $\tau$ ; (2) fold $h(\tau)$ to get $h(-\tau)$ ; (3) shift by $t$ to get $h(t-\tau)$ ; (4) multiply by $x(\tau)$ ; (5) integrate over the overlap; (6) repeat for all $t$ .

Convolution of $x(t)=e^{-t}u(t)$ and $y(t)=e^{-2t}u(t-3)$

Let $z(t) = x(t) * y(t)$ :

z(t) = \int_{-\infty}^{\infty} e^{-\tau}u(\tau)\; e^{-2(t-\tau)}u(t-\tau-3)\,d\tau

Limits:

$u(\tau) = 1$ only for $\tau \ge 0$ .
$u(t-\tau-3) = 1$ only for $\tau \le t-3$ .

So the product is non-zero only for $0 \le \tau \le t-3$ , which needs $t \ge 3$ .

Case 1: $t < 3$ – no overlap, so $z(t) = 0$ .

Case 2: $t \ge 3$ :

\begin{aligned} z(t) &= \int_{0}^{t-3} e^{-\tau}\,e^{-2t}\,e^{2\tau}\,d\tau \\ &= e^{-2t}\int_{0}^{t-3} e^{\tau}\,d\tau \\ &= e^{-2t}\left[e^{\tau}\right]_0^{t-3} \\ &= e^{-2t}\left(e^{t-3} - 1\right) \\ &= e^{-(t+3)} - e^{-2t} \end{aligned}

Check: at $t = 3$ , $z(3) = e^{-6} - e^{-6} = 0$ , so the output starts smoothly from zero. At $t = 4$ , $z(4) = e^{-7} - e^{-8} \approx 0.000576$ .

Answer:

z(t) = x(t)*y(t) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)

Shortcut check: $e^{-2t}u(t-3) = e^{-6}\,e^{-2(t-3)}u(t-3)$ , so it is $e^{-6}$ times $e^{-2t}u(t)$ delayed by 3. Since $e^{-t}u(t) * e^{-2t}u(t) = (e^{-t}-e^{-2t})u(t)$ , the result is $e^{-6}\left(e^{-(t-3)} - e^{-2(t-3)}\right)u(t-3) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)$ , the same answer.

2075 Bhadra · 4 marks

Determine whether the given system is linear or not. y(t) = A x(t) + B

Answer

A system is linear if it obeys superposition: for inputs $x_1, x_2$ and constants $a, b$ , the input $a x_1(t) + b x_2(t)$ must give the output $a y_1(t) + b y_2(t)$ .

Given: $y(t) = A\,x(t) + B$

Outputs for two separate inputs:

\begin{aligned} y_1(t) &= A x_1(t) + B \\ y_2(t) &= A x_2(t) + B \end{aligned}

Weighted sum of the outputs:

a y_1(t) + b y_2(t) = A\left[a x_1(t) + b x_2(t)\right] + (a+b)B

Output for the combined input $x_3(t) = a x_1(t) + b x_2(t)$ :

y_3(t) = A\left[a x_1(t) + b x_2(t)\right] + B

These two are equal only if $(a+b)B = B$ for all $a, b$ , i.e. only when $B = 0$ .

Zero-input test: for $x(t) = 0$ , $y(t) = B \ne 0$ . A linear system must give zero output for zero input, so the test fails.

Conclusion:

If $B \ne 0$ : the system is not linear. It is called an incrementally linear (affine) system: the change in output is linear in the change in input.
If $B = 0$ : $y(t) = A x(t)$ is linear.

2075 Bhadra · 3 marks

Write a short note on causal and non causal systems.

Answer

A causal system is one whose output at any time depends only on the present and past values of the input, never on future values. A non-causal system has an output that depends on at least one future input value.

Causal system

$y(t)$ depends on $x(\tau)$ only for $\tau \le t$ .
It does not respond before the input is applied (non-anticipative).
All physically realizable real-time systems are causal.
Examples: $y(t) = x(t) + 2x(t-1)$ ; an RC circuit; $y[n] = x[n] - x[n-1]$ .
LTI condition: $h(t) = 0$ for $t < 0$ (or $h[n] = 0$ for $n < 0$ ).

Non-causal system

The output uses future input, e.g. $y(t) = x(t+1)$ , $y[n] = x[n+1] + x[n]$ , $y(t) = x(-t)$ .
It cannot work in real time, but can be used when the whole signal is stored (image processing, offline audio smoothing).
An ideal low pass filter is non-causal because its $h(t)$ is a sinc that exists for $t < 0$ .

Anti-causal system: output depends only on future inputs, $h(t) = 0$ for $t > 0$ .

2075 Baisakh · 5 marks

Find output y(t) of LTI system if input x(t) = exp(−at) u(t) and impulse response h(t) = u(t).

Answer

For an LTI system the output is the convolution of input and impulse response:

y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

Given: $x(t) = e^{-at}u(t)$ (take $a > 0$ ), $h(t) = u(t)$ .

y(t) = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau

Limits: $u(\tau) = 1$ for $\tau \ge 0$ and $u(t-\tau) = 1$ for $\tau \le t$ . The overlap is $0 \le \tau \le t$ , which exists only for $t \ge 0$ .

For $t < 0$ : no overlap, $y(t) = 0$ .

For $t \ge 0$ :

\begin{aligned} y(t) &= \int_{0}^{t} e^{-a\tau}\,d\tau \\ &= \left[\frac{e^{-a\tau}}{-a}\right]_0^{t} \\ &= \frac{1}{a}\left(1 - e^{-at}\right) \end{aligned}

Answer:

y(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)

 y(t)
 1/a |        .-------------
     |     .-'
     |   .'
     |  /
     | /
   0 +/----------------------> t
     0

The output starts at 0 at $t = 0$ and rises to the final value $1/a$ as $t \to \infty$ . This is expected, because $h(t) = u(t)$ is an ideal integrator, so $y(t) = \int_{-\infty}^{t} x(\tau)\,d\tau$ , the running area under $e^{-at}$ , whose total area is $1/a$ .

2075 Baisakh · 4 marks

Derive the conditions for distortionless transmission for continuous-time LTI system.

Answer

Distortionless transmission means the output is an exact copy of the input in shape; it may only be scaled in amplitude and delayed in time.

Time-domain condition:

y(t) = K\,x(t - t_0)

where $K$ is a constant gain and $t_0 \ge 0$ is a constant delay.

Impulse response: put $x(t) = \delta(t)$ :

h(t) = K\,\delta(t - t_0)

Frequency response: take the Fourier transform of $y(t) = K x(t-t_0)$ using the time-shift property:

\begin{aligned} Y(\omega) &= K\,X(\omega)\,e^{-j\omega t_0} \\ H(\omega) &= \frac{Y(\omega)}{X(\omega)} = K\,e^{-j\omega t_0} \end{aligned}

Conditions:

Constant magnitude: $|H(\omega)| = K$ for all frequencies (at least over the band of the signal). Every frequency component is amplified equally, so there is no amplitude distortion.
Linear phase: $\theta(\omega) = \angle H(\omega) = -\omega t_0$ (a straight line through the origin with slope $-t_0$ ). Then every frequency component is delayed by the same time, so there is no phase distortion.

The delay is constant because the group delay is

t_g = -\frac{d\theta(\omega)}{d\omega} = t_0

 |H(w)|                 angle H(w)
  K  +--------------      |\
     |                    |  \  slope = -t0
     |                    |    \
   --+-------------> w  --+------\----> w

In practice, it is enough for these conditions to hold over the bandwidth of the input signal.

2073 Magh · 2+8 marks

What is LTI system? For LTI system, describe the properties: (a) linearity (b) stability (c) time invariance and (d) causality.

Answer

A Linear Time-Invariant (LTI) system is a system that is both linear (obeys superposition) and time invariant (its behaviour does not change with time). An LTI system is completely described by its impulse response $h(t)$ , and its output for any input is the convolution $y(t) = x(t) * h(t)$ .

(a) Linearity

A system is linear if it satisfies superposition = additivity + homogeneity:

a\,x_1(t) + b\,x_2(t) \;\rightarrow\; a\,y_1(t) + b\,y_2(t)

Additivity: $x_1 + x_2 \rightarrow y_1 + y_2$ .
Homogeneity (scaling): $a x \rightarrow a y$ .
A consequence: zero input gives zero output.
Examples: $y(t) = 3x(t)$ , $y(t) = \frac{dx}{dt}$ are linear; $y(t) = x^2(t)$ and $y(t) = 2x(t) + 1$ are not.
Convolution itself is a linear operation, so an LTI system is linear in its input.

(b) Stability

A system is BIBO stable (bounded-input bounded-output) if every bounded input $|x(t)| \le M_x < \infty$ gives a bounded output $|y(t)| \le M_y < \infty$ .

For an LTI system:

|y(t)| = \left|\int h(\tau)x(t-\tau)d\tau\right| \le M_x \int_{-\infty}^{\infty}|h(\tau)|\,d\tau

So the condition is that the impulse response is absolutely integrable:

\int_{-\infty}^{\infty} |h(t)|\,dt < \infty

$h(t) = e^{-2t}u(t)$ : area $= 1/2$ , stable.
$h(t) = u(t)$ (integrator): area infinite, unstable.

(c) Time invariance

A system is time invariant if a time shift of the input causes the same time shift of the output:

x(t) \rightarrow y(t) \;\Rightarrow\; x(t - t_0) \rightarrow y(t - t_0)

Test: find the output for the delayed input, then compare with $y(t-t_0)$ .
$y(t) = x(t-2)$ is time invariant.
$y(t) = t\,x(t)$ is time varying, because the delayed input gives $t\,x(t-t_0)$ but $y(t-t_0) = (t-t_0)x(t-t_0)$ .
Systems with constant coefficients are time invariant.

(d) Causality

A system is causal if its output at time $t$ depends only on present and past inputs ( $\tau \le t$ ), not on future inputs.

For an LTI system, $y(t) = \int h(\tau)x(t-\tau)d\tau$ . To avoid using future values $x(t-\tau)$ with $\tau < 0$ :

h(t) = 0 \quad \text{for } t < 0

Then $y(t) = \int_{0}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$ .

$h(t) = e^{-t}u(t)$ is causal.
$h(t) = e^{-|t|}$ is non-causal (it is stable, though).

Property	General condition	LTI condition
Linearity	Superposition holds	Always (convolution)
Time invariance	Shifted input gives shifted output	Always
Causality	No future inputs used	$h(t)=0$ , $t<0$
Stability	Bounded in, bounded out	$\int \lvert h(t)\rvert dt < \infty$

2073 Bhadra · 4+2 marks

Derive Formula to calculate the impulse response of continuous time ideal low pass filter. Is this system practically realizable or not.

Answer

Impulse response of an ideal low pass filter

An ideal low pass filter (LPF) passes all frequencies below the cutoff $\omega_c$ without distortion and completely blocks higher frequencies:

H(\omega) = \begin{cases} K\,e^{-j\omega t_0}, & |\omega| < \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

(constant gain $K$ and linear phase, i.e. distortionless in the pass band).

 |H(w)|
   K  +-----------+
      |           |
 -----+-----+-----+-------> w
    -wc     0     wc

The impulse response is the inverse Fourier transform:

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} K\,e^{-j\omega t_0}\,e^{j\omega t}\,d\omega \\ &= \frac{K}{2\pi}\left[\frac{e^{j\omega(t-t_0)}}{j(t-t_0)}\right]_{-\omega_c}^{\omega_c} \\ &= \frac{K}{2\pi}\cdot\frac{2j\sin\omega_c(t-t_0)}{j(t-t_0)} \\ &= \frac{K\sin\omega_c(t-t_0)}{\pi(t-t_0)} \\ &= \frac{K\omega_c}{\pi}\,\mathrm{sinc}\left[\omega_c(t-t_0)\right] \end{aligned}

where $\mathrm{sinc}(x) = \sin x / x$ . The peak value $K\omega_c/\pi$ occurs at $t = t_0$ , and zeros occur at $t = t_0 \pm n\pi/\omega_c$ .

            h(t)
             /\    peak K*wc/pi at t0
   .  ..   /  \   ..  .
 --'--''--/----\--''--'--> t
          |  t0 |
   tails extend to t -> -inf

Is it practically realizable?

No. A physically realizable system must be causal, i.e. $h(t) = 0$ for $t < 0$ . The sinc response extends from $t = -\infty$ to $+\infty$ , so it is non-zero for $t < 0$ for any finite delay $t_0$ : the filter would respond before the impulse is applied. Hence the ideal LPF is non-causal and not realizable. (The Paley–Wiener criterion also fails, because $|H(\omega)|$ is exactly zero over a band.) Practical filters (Butterworth, Chebyshev) only approximate it, using a large delay and truncating the tails.

2073 Bhadra · 3 marks

Write a short note on invertibility of LTI system.

Answer

A system is invertible if distinct inputs always give distinct outputs, so that the input can be recovered exactly from the output. The system that recovers it is called the inverse system.

x(t) --> [ h(t) ] --> y(t) --> [ h_inv(t) ] --> x(t)

Condition for an LTI system: the system $h(t)$ is invertible if there is an LTI system $h_{inv}(t)$ such that the cascade gives the identity system:

h(t) * h_{inv}(t) = \delta(t)

In the frequency domain: $H(\omega)\,H_{inv}(\omega) = 1$ , so $H_{inv}(\omega) = 1/H(\omega)$ . This needs $H(\omega) \ne 0$ at every frequency.

Examples:

$y(t) = x(t - t_0)$ , $h(t) = \delta(t-t_0)$ : inverse $h_{inv}(t) = \delta(t+t_0)$ .
Integrator $h(t) = u(t)$ : inverse is the differentiator, $y(t) = \frac{dx}{dt}$ .
$y[n] = x[n] - x[n-1]$ : inverse is the accumulator $y[n] = \sum_{k=-\infty}^{n} x[k]$ .
Not invertible: $y(t) = 0$ , $y(t) = x^2(t)$ (sign is lost), an ideal LPF (high frequencies are lost).

Use: channel equalization in communication, deconvolution, decoding.

2072 Asoj · 2+6 marks

What is convolution? Obtain the expression for convolution integral.

Answer

Convolution

Convolution is a mathematical operation that combines two signals to produce a third. In signals and systems it gives the output $y(t)$ of an LTI system as the input $x(t)$ "convolved" with the impulse response $h(t)$ :

y(t) = x(t) * h(t)

It works as a weighted, shifted sum: each past value of the input is weighted by the system's impulse response and all contributions are added.

Expression for the convolution integral

Step 1 – Input as a sum of impulses. Approximate $x(t)$ by narrow pulses of width $\Delta$ :

x(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_\Delta(t-k\Delta)\,\Delta

where $\delta_\Delta(t)$ is a pulse of width $\Delta$ and height $1/\Delta$ (unit area).

Step 2 – Response to one pulse. Let $h_\Delta(t)$ be the response to $\delta_\Delta(t)$ .

Time invariance: $\delta_\Delta(t-k\Delta) \rightarrow h_\Delta(t-k\Delta)$ .
Homogeneity: $x(k\Delta)\Delta\,\delta_\Delta(t-k\Delta) \rightarrow x(k\Delta)\Delta\,h_\Delta(t-k\Delta)$ .
Additivity:

y(t) \approx \sum_{k=-\infty}^{\infty} x(k\Delta)\,h_\Delta(t-k\Delta)\,\Delta

Step 3 – Limit $\Delta \to 0$ . Then $\delta_\Delta \to \delta(t)$ , $h_\Delta \to h(t)$ , $k\Delta \to \tau$ , $\Delta \to d\tau$ and the sum becomes an integral:

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

This is the convolution integral. Substituting $\lambda = t-\tau$ gives the equivalent form $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$ .

For causal system and causal input ( $h(t) = 0$ and $x(t) = 0$ for $t < 0$ ):

y(t) = \int_{0}^{t} x(\tau)\,h(t-\tau)\,d\tau

Procedure (graphical): fold $h(\tau)$ to $h(-\tau)$ , shift it by $t$ , multiply by $x(\tau)$ , and find the area of the product; repeat for each $t$ .

Properties: commutative ( $x*h = h*x$ ), associative ( $(x*h_1)*h_2 = x*(h_1*h_2)$ , used for cascade), distributive ( $x*(h_1+h_2) = x*h_1 + x*h_2$ , used for parallel connection), and $x(t)*\delta(t-t_0) = x(t-t_0)$ .

Example: $e^{-at}u(t) * u(t) = \int_0^t e^{-a\tau}d\tau = \frac{1}{a}(1-e^{-at})u(t)$ .

2071 Magh · 7 marks

Find convolution between two signals x(t) = e^(0.5t) for 0 < t < 5, 0 otherwise and h(t) = 1 for 1 < t < 3, 0 otherwise.

Answer

Use the convolution integral, sliding the pulse $h$ across the fixed input $x$ :

y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau

Given:

$x(\tau) = e^{0.5\tau}$ for $0 < \tau < 5$ .
$h(t-\tau) = 1$ when $1 < t-\tau < 3$ , i.e. $t-3 < \tau < t-1$ .

So the shifted window is $(t-3,\ t-1)$ , of width 2, and it must overlap $(0,\ 5)$ .

x(tau):        |0=========5|
window:  (t-3 ----- t-1)  slides right as t increases

Useful integral: $\int_{a}^{b} e^{0.5\tau}d\tau = 2\left(e^{0.5b} - e^{0.5a}\right)$ .

Case 1: $t < 1$ ( $t-1 < 0$ ): no overlap, $y(t) = 0$ .

Case 2: $1 < t < 3$ (partial overlap, window enters): limits $0$ to $t-1$ .

y(t) = \int_0^{t-1} e^{0.5\tau}d\tau = 2\left(e^{0.5(t-1)} - 1\right)

Case 3: $3 < t < 6$ (window fully inside): limits $t-3$ to $t-1$ .

\begin{aligned} y(t) &= 2\left(e^{0.5(t-1)} - e^{0.5(t-3)}\right) \\ &= 2\left(1 - e^{-1}\right)e^{0.5(t-1)} \approx 1.2642\,e^{0.5(t-1)} \end{aligned}

Case 4: $6 < t < 8$ (window leaving): limits $t-3$ to $5$ .

y(t) = 2\left(e^{2.5} - e^{0.5(t-3)}\right)

Case 5: $t > 8$ ( $t-3 > 5$ ): no overlap, $y(t) = 0$ .

Answer:

y(t) = \begin{cases} 0, & t < 1 \\ 2\left(e^{0.5(t-1)} - 1\right), & 1 \le t < 3 \\ 2\left(e^{0.5(t-1)} - e^{0.5(t-3)}\right), & 3 \le t < 6 \\ 2\left(e^{2.5} - e^{0.5(t-3)}\right), & 6 \le t < 8 \\ 0, & t \ge 8 \end{cases}

Key values (continuity check):

$t$	$y(t)$
1	0
3	$2(e - 1) = 3.437$
6	$2(e^{2.5} - e^{1.5}) = 15.402$
8	0

The pieces meet at $t = 3$ and $t = 6$ . The output rises from 0 at $t=1$ , peaks at about 15.40 at $t = 6$ , and falls to 0 at $t = 8$ . The total duration is $5 + 2 = 7$ (from $0+1$ to $5+3$ ), as expected for convolution.

 y(t)
15.4|                 *
    |              .-' \
    |          .--'     \
 3.4|      .--*          \
    |   .-'               \
  0 +--*-----+-------------*--> t
       1     3        6    8

2071 Magh · 4+4 marks

Derive the expression for impulse response and step response of ideal low pass filter.

Answer

Impulse response of the ideal LPF

The ideal LPF has constant gain and linear phase in the pass band and zero gain outside:

H(\omega) = \begin{cases} e^{-j\omega t_0}, & |\omega| \le \omega_c \\ 0, & |\omega| > \omega_c \end{cases}

(unit gain taken; for gain $K$ , multiply the results by $K$ ).

Inverse Fourier transform:

\begin{aligned} h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} H(\omega)e^{j\omega t}d\omega = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega \\ &= \frac{1}{2\pi}\cdot\frac{e^{j\omega_c(t-t_0)} - e^{-j\omega_c(t-t_0)}}{j(t-t_0)} \\ &= \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)} = \frac{\omega_c}{\pi}\,\mathrm{sinc}\left[\omega_c(t-t_0)\right] \end{aligned}

Peak value $\omega_c/\pi$ at $t = t_0$ .
Zero crossings at $t = t_0 + n\pi/\omega_c$ , $n = \pm1, \pm2, \dots$
Main lobe width $2\pi/\omega_c$ : a wider bandwidth gives a narrower, taller pulse.
$h(t) \ne 0$ for $t < 0$ , so the filter is non-causal (not physically realizable).

           h(t)
            /\   wc/pi
  .  .-.   /  \   .-.  .
 -'--' '--/----\--' '--'--> t
             t0

Step response of the ideal LPF

The step response is the running integral of the impulse response:

s(t) = \int_{-\infty}^{t} h(\lambda)\,d\lambda = \int_{-\infty}^{t} \frac{\sin\omega_c(\lambda-t_0)}{\pi(\lambda-t_0)}\,d\lambda

Let $x = \omega_c(\lambda - t_0)$ , so $d\lambda = dx/\omega_c$ and $\lambda - t_0 = x/\omega_c$ :

\begin{aligned} s(t) &= \frac{1}{\pi}\int_{-\infty}^{\omega_c(t-t_0)} \frac{\sin x}{x}\,dx \\ &= \frac{1}{\pi}\int_{-\infty}^{0} \frac{\sin x}{x}\,dx + \frac{1}{\pi}\int_{0}^{\omega_c(t-t_0)} \frac{\sin x}{x}\,dx \end{aligned}

Using $\int_{-\infty}^{0} \frac{\sin x}{x}dx = \frac{\pi}{2}$ and the sine integral $\mathrm{Si}(y) = \int_0^{y}\frac{\sin x}{x}dx$ :

s(t) = \frac{1}{2} + \frac{1}{\pi}\,\mathrm{Si}\left[\omega_c(t-t_0)\right]

Features:

$s(t_0) = 1/2$ ; $s(-\infty) = 0$ ; $s(\infty) = 1$ (since $\mathrm{Si}(\pm\infty) = \pm\pi/2$ ).
Rise time: the slope at $t_0$ is $h(t_0) = \omega_c/\pi$ , so $t_r \approx \pi/\omega_c = 1/(2B)$ where $B = \omega_c/2\pi$ Hz. Rise time is inversely proportional to bandwidth.
Overshoot and ringing: the response overshoots by about 9% (peak about 1.09) and oscillates around 1. This is the Gibbs phenomenon; it does not shrink as $\omega_c$ increases.
The response starts before $t = 0$ , again showing the filter is non-causal.

 s(t)
 1.09|          _
 1.0 |   - - - / \_/~~~~~~~
 0.5 |       /
     |~~\_  /
   0 +----------------------> t
            t0

2071 Magh · 6 marks

For the LTI system, describe following properties: (a) linearity (b) causality (c) stability (d) time invariance.

Answer

An LTI system is linear and time invariant; it is fully described by its impulse response $h(t)$ , and $y(t) = x(t) * h(t)$ .

(a) Linearity

A system is linear if it obeys superposition:

a\,x_1(t) + b\,x_2(t) \rightarrow a\,y_1(t) + b\,y_2(t)

It combines additivity ( $x_1 + x_2 \rightarrow y_1 + y_2$ ) and homogeneity ( $a x \rightarrow a y$ ). Zero input must give zero output.

Linear: $y(t) = 5x(t)$ , $y(t) = \int_{-\infty}^{t} x(\tau)d\tau$ .
Non-linear: $y(t) = x^2(t)$ , $y(t) = x(t) + 3$ .

(b) Causality

The output at any time depends only on the present and past inputs. For an LTI system:

h(t) = 0 \quad \text{for } t < 0

so that $y(t) = \int_0^{\infty} h(\tau)x(t-\tau)d\tau$ uses only $x$ up to time $t$ .

Causal: $h(t) = e^{-3t}u(t)$ .
Non-causal: $h(t) = \delta(t+1)$ , i.e. $y(t) = x(t+1)$ .

(c) Stability

\int_{-\infty}^{\infty} |h(t)|\,dt < \infty

Stable: $h(t) = e^{-t}u(t)$ (area 1).
Unstable: $h(t) = u(t)$ , $h(t) = e^{t}u(t)$ .

(d) Time invariance

A time shift in the input gives the same time shift in the output:

x(t-t_0) \rightarrow y(t-t_0)

The system's characteristics do not change with time (constant parameters).

Time invariant: $y(t) = x(t-1)$ , $y(t) = \frac{dx}{dt}$ .
Time varying: $y(t) = t\,x(t)$ , $y(t) = x(2t)$ .

Property	LTI test on $h(t)$
Linearity	Holds for every convolution system
Causality	$h(t) = 0$ for $t < 0$
Stability	$\int \lvert h(t)\rvert dt$ finite
Time invariance	Holds for every convolution system

2070 Magh · 8+2 marks

The impulse responses of two LTI systems are given by h₁(t) = e^(−t/2)u(t) and h₂(t) = u(t) − u(t−5). Determine the equivalent impulse response if these two systems are connected in cascade. Also sketch the graph of equivalent impulse response.

Answer

For two LTI systems in cascade, the equivalent impulse response is the convolution of the individual impulse responses (associative property):

h(t) = h_1(t) * h_2(t) = \int_{-\infty}^{\infty} h_1(\tau)\,h_2(t-\tau)\,d\tau

x(t) -->[ h1(t) ]-->[ h2(t) ]--> y(t)
   ==  x(t) -->[ h1(t)*h2(t) ]--> y(t)

Given: $h_1(t) = e^{-t/2}u(t)$ ; $h_2(t) = u(t) - u(t-5)$ , a unit pulse on $0 \le t < 5$ .

$h_2(t-\tau) = 1$ when $0 \le t-\tau < 5$ , i.e. $t-5 < \tau \le t$ . Also $h_1(\tau) \ne 0$ only for $\tau \ge 0$ .

Useful integral: $\int_a^b e^{-\tau/2}d\tau = 2\left(e^{-a/2} - e^{-b/2}\right)$ .

Case 1: $t < 0$ – no overlap: $h(t) = 0$ .

Case 2: $0 \le t < 5$ – limits $0$ to $t$ :

h(t) = \int_0^{t} e^{-\tau/2}d\tau = 2\left(1 - e^{-t/2}\right)

Case 3: $t \ge 5$ – limits $t-5$ to $t$ :

\begin{aligned} h(t) &= 2\left(e^{-(t-5)/2} - e^{-t/2}\right) \\ &= 2\left(e^{2.5} - 1\right)e^{-t/2} \approx 22.365\,e^{-t/2} \end{aligned}

Answer:

h(t) = \begin{cases} 0, & t < 0 \\ 2\left(1 - e^{-t/2}\right), & 0 \le t < 5 \\ 2\left(e^{2.5} - 1\right)e^{-t/2}, & t \ge 5 \end{cases}

Compactly: $h(t) = 2\left(1 - e^{-t/2}\right)u(t) - 2\left(1 - e^{-(t-5)/2}\right)u(t-5)$ .

Values for the sketch:

$t$	0	1	2	5	7	10
$h(t)$	0	0.787	1.264	1.836	0.675	0.151

Peak: $h(5) = 2(1 - e^{-2.5}) = 1.836$ , which is continuous from both sides.

Sketch

 h(t)
1.84|            *
    |         .-' \
    |      .-'     \
    |    .'         `.
    |  .'             `-._
    | /                   `--.__
  0 +-------------+---------------> t
    0             5     10

It rises like a charging capacitor ( $2(1-e^{-t/2})$ ) from 0 to 1.836 over $0 \le t \le 5$ , then decays exponentially to zero with time constant 2 s for $t > 5$ .

2070 Bhadra · 5 marks

Let x(t) be the input to an LTI system with unit impulse response h(t), where x(t) = e^(−at)u(t), a > 0 and h(t) = u(t). Verify commutative law of LTI system.

Answer

The commutative law of convolution states $x(t) * h(t) = h(t) * x(t)$ : the output of an LTI system is the same whether we convolve the input with the impulse response or the other way round.

Given: $x(t) = e^{-at}u(t)$ , $a > 0$ ; $h(t) = u(t)$ .

LHS: $y_1(t) = x(t) * h(t)$

y_1(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$ ):

y_1(t) = \int_0^{t} e^{-a\tau}d\tau = \left[\frac{e^{-a\tau}}{-a}\right]_0^t = \frac{1}{a}\left(1 - e^{-at}\right)u(t)

RHS: $y_2(t) = h(t) * x(t)$

y_2(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau = \int_{-\infty}^{\infty} u(\tau)\,e^{-a(t-\tau)}u(t-\tau)\,d\tau

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$ ):

\begin{aligned} y_2(t) &= e^{-at}\int_0^{t} e^{a\tau}d\tau \\ &= e^{-at}\cdot\frac{e^{at} - 1}{a} \\ &= \frac{1}{a}\left(1 - e^{-at}\right)u(t) \end{aligned}

Result

y_1(t) = y_2(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)

so $x(t) * h(t) = h(t) * x(t)$ and the commutative law is verified.

Meaning: an input $e^{-at}u(t)$ applied to an integrator ( $h = u(t)$ ) gives the same output as a unit step applied to a system with impulse response $e^{-at}u(t)$ . In general the roles of input and impulse response can be interchanged.

2070 Bhadra · 5 marks

What is distortionless transmission? Derive the expression for unit step response of ideal low pass filter.

Answer

Distortionless transmission

Transmission is distortionless when the output has exactly the same shape as the input, only scaled by a constant $K$ and delayed by a constant $t_0$ :

y(t) = K\,x(t-t_0) \;\Rightarrow\; H(\omega) = K\,e^{-j\omega t_0}

So the system needs (1) constant magnitude $|H(\omega)| = K$ and (2) linear phase $\angle H(\omega) = -\omega t_0$ over the signal bandwidth.

Unit step response of the ideal LPF

Ideal LPF (unit gain, delay $t_0$ , cutoff $\omega_c$ ):

H(\omega) = e^{-j\omega t_0}\ \text{for } |\omega| < \omega_c,\quad 0 \text{ otherwise}

Its impulse response (inverse Fourier transform) is

h(t) = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega = \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}

The step response is the integral of $h(t)$ :

s(t) = \int_{-\infty}^{t} \frac{\sin\omega_c(\lambda-t_0)}{\pi(\lambda-t_0)}\,d\lambda

Put $x = \omega_c(\lambda - t_0)$ :

\begin{aligned} s(t) &= \frac{1}{\pi}\int_{-\infty}^{\omega_c(t-t_0)}\frac{\sin x}{x}dx \\ &= \frac{1}{\pi}\left[\frac{\pi}{2} + \mathrm{Si}\left(\omega_c(t-t_0)\right)\right] \\ &= \frac{1}{2} + \frac{1}{\pi}\mathrm{Si}\left[\omega_c(t-t_0)\right] \end{aligned}

where $\mathrm{Si}(y) = \int_0^{y}\frac{\sin x}{x}dx$ is the sine integral.

$s(t_0) = 0.5$ , $s(\infty) = 1$ , $s(-\infty) = 0$ .
Rise time $\approx \pi/\omega_c$ , inversely proportional to bandwidth.
About 9% overshoot with ringing (Gibbs phenomenon), and response before $t = 0$ (non-causal).

2069 Bhadra · 8 marks

A LTI system has input x(t) = e^(−t)u(t) and impulse response h(t) = e^(t)u(−t). Find output y(t) of the system using Fourier transform of x(t) and h(t).

Answer

For an LTI system, convolution in time becomes multiplication in frequency:

y(t) = x(t) * h(t) \;\Longleftrightarrow\; Y(\omega) = X(\omega)\,H(\omega)

Step 1: Fourier transform of the input

\begin{aligned} X(\omega) &= \int_{0}^{\infty} e^{-t}e^{-j\omega t}dt = \left[\frac{e^{-(1+j\omega)t}}{-(1+j\omega)}\right]_0^{\infty} \\ &= \frac{1}{1 + j\omega} \end{aligned}

Step 2: Fourier transform of the impulse response

$h(t) = e^{t}u(-t)$ is non-zero only for $t < 0$ :

\begin{aligned} H(\omega) &= \int_{-\infty}^{0} e^{t}e^{-j\omega t}dt = \left[\frac{e^{(1-j\omega)t}}{1-j\omega}\right]_{-\infty}^{0} \\ &= \frac{1}{1 - j\omega} \end{aligned}

Step 3: Output spectrum

Y(\omega) = \frac{1}{(1+j\omega)(1-j\omega)} = \frac{1}{1 + \omega^2}

Step 4: Inverse transform (partial fractions)

\frac{1}{(1+j\omega)(1-j\omega)} = \frac{A}{1+j\omega} + \frac{B}{1-j\omega}

Multiply out: $1 = A(1-j\omega) + B(1+j\omega)$ . Comparing terms: $A + B = 1$ and $-A + B = 0$ , so $A = B = \tfrac{1}{2}$ .

Y(\omega) = \frac{1}{2}\cdot\frac{1}{1+j\omega} + \frac{1}{2}\cdot\frac{1}{1-j\omega}

Using the pairs from Steps 1 and 2:

$\frac{1}{1+j\omega} \leftrightarrow e^{-t}u(t)$
$\frac{1}{1-j\omega} \leftrightarrow e^{t}u(-t)$

y(t) = \frac{1}{2}e^{-t}u(t) + \frac{1}{2}e^{t}u(-t)

Answer:

y(t) = \frac{1}{2}e^{-|t|}, \quad -\infty < t < \infty

Check: the standard pair $e^{-a|t|} \leftrightarrow \frac{2a}{a^2+\omega^2}$ with $a = 1$ gives $\frac{1}{2}e^{-|t|} \leftrightarrow \frac{1}{1+\omega^2}$ . Also $y(0) = \frac{1}{2\pi}\int\frac{d\omega}{1+\omega^2} = \frac{1}{2\pi}\cdot\pi = 0.5$ .

          y(t)
          0.5
          /\
       .-'  '-.
 ___.-'        '-.___
 ---------+----------> t
          0

The output is two-sided because $h(t)$ is anti-causal (non-zero for $t < 0$ ).

2069 Bhadra · 5 marks

Write about the following properties of continuous time system: (a) Linearity (b) Causality (c) Memory (d) Stability (e) Time invariance.

Answer

Continuous-time systems are classified by the following basic properties.

(a) Linearity

A system is linear if it obeys superposition (additivity + homogeneity): $a x_1(t) + b x_2(t) \rightarrow a y_1(t) + b y_2(t)$ .

Linear: $y(t) = 2x(t)$ , $y(t) = \frac{dx}{dt}$ .
Non-linear: $y(t) = x^2(t)$ , $y(t) = x(t) + 1$ .

(b) Causality

Output at time $t$ depends only on present and past inputs, not future ones. It is non-anticipative and can work in real time.

Causal: $y(t) = x(t) + x(t-2)$ .
Non-causal: $y(t) = x(t+1)$ . For LTI: causal if $h(t) = 0$ for $t < 0$ .

(c) Memory

A memoryless (static) system's output depends only on the input at the same instant. A system with memory (dynamic) uses past (or future) inputs.

Memoryless: $y(t) = 5x(t)$ , a resistor $v = Ri$ .
With memory: $y(t) = \int_{-\infty}^{t} x(\tau)d\tau$ , a capacitor. For LTI: memoryless only if $h(t) = K\delta(t)$ .

(d) Stability

BIBO stable: every bounded input gives a bounded output.

Stable: $y(t) = e^{x(t)}$ , $h(t) = e^{-t}u(t)$ .
Unstable: $y(t) = t\,x(t)$ , an ideal integrator. For LTI: $\int|h(t)|dt < \infty$ .

(e) Time invariance

A time shift in the input causes the same shift in the output: $x(t-t_0) \rightarrow y(t-t_0)$ .

Time invariant: $y(t) = x(t-3)$ .
Time varying: $y(t) = t\,x(t)$ , $y(t) = x(2t)$ .

2069 Bhadra · 5 marks

Derive the expression for impulse response and step response of first order continuous time system described by the differential equation τ dy(t)/dt + y(t) = x(t).

Answer

Given: a first-order system with time constant $\tau$ :

\tau\frac{dy(t)}{dt} + y(t) = x(t)

(e.g. an RC low pass circuit with $\tau = RC$ ). Assume the system is initially at rest.

Impulse response

Take the Fourier transform, using $\frac{d}{dt} \leftrightarrow j\omega$ :

\begin{aligned} (j\omega\tau + 1)\,Y(\omega) &= X(\omega) \\ H(\omega) = \frac{Y(\omega)}{X(\omega)} &= \frac{1}{1 + j\omega\tau} = \frac{1/\tau}{1/\tau + j\omega} \end{aligned}

Using the pair $e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$ with $a = 1/\tau$ :

h(t) = \frac{1}{\tau}e^{-t/\tau}u(t)

Check: for $t > 0$ , $\tau h' + h = -\frac{1}{\tau}e^{-t/\tau} + \frac{1}{\tau}e^{-t/\tau} = 0$ , and the jump $h(0^+) = 1/\tau$ makes $\tau\,dh/dt$ contain $\delta(t)$ , matching the input.

Step response

\begin{aligned} s(t) &= \int_{-\infty}^{t} h(\lambda)\,d\lambda = \int_0^{t}\frac{1}{\tau}e^{-\lambda/\tau}d\lambda \\ &= \left[-e^{-\lambda/\tau}\right]_0^{t} \\ &= \left(1 - e^{-t/\tau}\right)u(t) \end{aligned}

Features:

$s(\tau) = 1 - e^{-1} = 0.632$ : the output reaches 63.2% of the final value in one time constant.
It reaches about 98% after $4\tau$ ; final value is 1.
Smaller $\tau$ means a faster response (wider bandwidth $1/\tau$ rad/s).

 h(t)                      s(t)
1/tau|\                    1 |      .--------
     | \                0.632|   .-'
     |  `-.                  |  /
     |     `---___           | /
   0 +--------------> t    0 +/-------------> t
     0  tau                  0  tau

2069 Bhadra · 5 marks

If the impulse response of continuous time linear time invariant system is h(t) = u(t) − u(t−3) and input to the system is x(t) = u(t+4) − u(t), determine the output y(t) of the system.

Answer

The output is the convolution $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$ .

Given:

$h(t) = u(t) - u(t-3)$ : unit pulse on $0 \le t < 3$ (width 3).
$x(t) = u(t+4) - u(t)$ : unit pulse on $-4 \le t < 0$ (width 4).

The output lasts from $0 + (-4) = -4$ to $3 + 0 = 3$ , and convolving two unequal rectangles gives a trapezoid with flat top height = smaller width = 3.

Limits: $x(t-\tau) = 1$ when $-4 \le t-\tau < 0$ , i.e. $t < \tau \le t+4$ . Combine with $0 \le \tau < 3$ from $h$ .

Case 1: $t < -4$ ( $t+4 < 0$ ): no overlap, $y(t) = 0$ .

Case 2: $-4 \le t < -1$ (partial overlap, $0 \le \tau \le t+4$ ):

y(t) = \int_0^{t+4} d\tau = t + 4

Case 3: $-1 \le t < 0$ (window $(t,\ t+4)$ covers all of $[0,3]$ ):

y(t) = \int_0^{3} d\tau = 3

Case 4: $0 \le t < 3$ (partial overlap, $t \le \tau \le 3$ ):

y(t) = \int_t^{3} d\tau = 3 - t

Case 5: $t \ge 3$ : no overlap, $y(t) = 0$ .

Answer:

y(t) = \begin{cases} 0, & t < -4 \\ t + 4, & -4 \le t < -1 \\ 3, & -1 \le t < 0 \\ 3 - t, & 0 \le t < 3 \\ 0, & t \ge 3 \end{cases}

Equivalently $y(t) = r(t+4) - r(t+1) - r(t) + r(t-3)$ , where $r(t) = t\,u(t)$ is the ramp.

Check: area of $y$ = (area of $h$ ) × (area of $x$ ) = $3 \times 4 = 12$ ; trapezoid area $= \frac{1}{2}(7 + 1)\times 3 = 12$ .

 y(t)
   3 |        +----+
     |       /|    |\
     |      / |    | \
     |     /  |    |  \
   0 +----+---+----+---+----> t
         -4  -1    0   3

2083 Bhadra (new course) · 5 marks

Define LTI system. Explain causality and stability property of continuous time LTI system.

Answer

A Linear Time-Invariant (LTI) system is a system that obeys superposition (linear) and whose response to a delayed input is the same response delayed (time invariant). Its output for any input is given by convolution with its impulse response:

y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau

Causality

A system is causal if its output at time $t$ depends only on inputs at times $\le t$ .

In the convolution, the term $x(t-\tau)$ for $\tau < 0$ is a future input value. For it never to be used, the weight $h(\tau)$ must be zero there:

h(t) = 0 \quad \text{for } t < 0

Then $y(t) = \int_0^{\infty} h(\tau)\,x(t-\tau)\,d\tau$ .

Causal: $h(t) = e^{-2t}u(t)$ , $h(t) = u(t) - u(t-1)$ .
Non-causal: $h(t) = e^{-|t|}$ , $h(t) = \delta(t+2)$ .
Physically realizable real-time systems must be causal.

Stability

A system is BIBO stable if every bounded input, $|x(t)| \le M_x$ , gives a bounded output.

|y(t)| = \left|\int h(\tau)x(t-\tau)d\tau\right| \le \int|h(\tau)|\,|x(t-\tau)|\,d\tau \le M_x\int_{-\infty}^{\infty}|h(\tau)|\,d\tau

So the output is bounded if

\int_{-\infty}^{\infty} |h(t)|\,dt < \infty

(the impulse response is absolutely integrable). This is also necessary: if the integral is infinite, the bounded input $x(t) = \mathrm{sgn}[h(-t)]$ gives $y(0) = \int|h(\tau)|d\tau = \infty$ .

Stable: $h(t) = e^{-2t}u(t)$ , since $\int_0^\infty e^{-2t}dt = 0.5$ .
Unstable: $h(t) = u(t)$ (integrator), $h(t) = e^{2t}u(t)$ .

2082 Bhadra (new course) · 2+6 marks

Define frequency response of a LTI system. Derive the impulse response of an ideal low pass and high pass filter.

Answer

Frequency response

The frequency response $H(\omega)$ of an LTI system is the Fourier transform of its impulse response. It tells how the system changes the magnitude and phase of a sinusoid of each frequency:

H(\omega) = \int_{-\infty}^{\infty} h(t)e^{-j\omega t}dt = \frac{Y(\omega)}{X(\omega)} = |H(\omega)|\,e^{j\theta(\omega)}

If the input is $e^{j\omega t}$ , the output is $H(\omega)e^{j\omega t}$ . $|H(\omega)|$ is the magnitude (gain) response and $\theta(\omega)$ is the phase response.

Impulse response of an ideal low pass filter

Pass band $|\omega| < \omega_c$ , unit gain, linear phase (delay $t_0$ ):

H_{LP}(\omega) = \begin{cases} e^{-j\omega t_0}, & |\omega| < \omega_c \\ 0, & \text{otherwise} \end{cases}

\begin{aligned} h_{LP}(t) &= \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega(t-t_0)}d\omega \\ &= \frac{1}{2\pi}\cdot\frac{2j\sin\omega_c(t-t_0)}{j(t-t_0)} \\ &= \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)} = \frac{\omega_c}{\pi}\mathrm{sinc}\left[\omega_c(t-t_0)\right] \end{aligned}

A sinc pulse centred at $t_0$ with peak $\omega_c/\pi$ .

Impulse response of an ideal high pass filter

The ideal HPF blocks $|\omega| < \omega_c$ and passes everything above. It is the all-pass (delay) system minus the LPF:

H_{HP}(\omega) = e^{-j\omega t_0} - H_{LP}(\omega) = \begin{cases} 0, & |\omega| < \omega_c \\ e^{-j\omega t_0}, & |\omega| > \omega_c \end{cases}

 |H_LP|              |H_HP|
 1 +----+          1 ---+      +---
   |    |               |      |
 --+----+----> w   -----+--+---+---> w
 -wc  0  wc           -wc  0   wc

Since $e^{-j\omega t_0} \leftrightarrow \delta(t-t_0)$ :

h_{HP}(t) = \delta(t - t_0) - \frac{\sin\omega_c(t-t_0)}{\pi(t-t_0)}

Remarks:

Both impulse responses are non-zero for $t < 0$ (sinc tails), so both ideal filters are non-causal and cannot be built exactly; practical filters approximate them.
With $t_0 = 0$ : $h_{LP}(t) = \frac{\sin\omega_c t}{\pi t}$ and $h_{HP}(t) = \delta(t) - \frac{\sin\omega_c t}{\pi t}$ .

Past questions and answers

1.Derive the convolution integral for continuous time LTI system.

Step 1: Represent the input with narrow pulses

Step 2: Response to one pulse

Step 3: Take the limit Δ→0\Delta \to 0Δ→0

Meaning and evaluation

2.Derive the expression for impulse response of ideal low pass filter (with cutoff frequency ω_c) and discuss.

Derivation

Discussion

3.Evaluate the step response of RC filter.

Impulse response

Step response

Discussion

4.Describe the frequency response (transfer function) of a continuous time LTI system.

Eigenfunction property

Transfer function

Magnitude and phase

5.Define linear time invariant system. Explain causality and time variance properties of continuous time system.

LTI system

Causality

Time variance

6.Derive the transfer function of a RC low pass filter and plot the magnitude and phase.

Transfer function

Magnitude and phase

7.Find the output y(t) of LTI system if the impulse response h(t) = e^(−t)u(t−3) and input x(t) = u(t−2).

Step 1: Write both functions in τ\tauτ

Step 2: Range of ttt

Step 3: Evaluate for t>5t > 5t>5

Result

8.Derive the expression for impulse response of RC filter and plot the magnitude and phase response.

Impulse response

Magnitude and phase response

9.Find the convolution between the signals x(t) = 1 for −1 < t < 1, 0 otherwise and h(t) = 1 for −1 < t < 1, 0 otherwise.

Case 1: t<−2t < -2t<−2

Case 2: −2≤t<0-2 \le t < 0−2≤t<0 (partial overlap from left)

Case 3: 0≤t<20 \le t < 20≤t<2 (partial overlap from right)

Case 4: t≥2t \ge 2t≥2

Result

10.What are the properties of system? Determine whether the given system is time-variant or not. y(t) = sin[x(t)]

Properties of systems

Is y(t)=sin⁡[x(t)]y(t) = \sin[x(t)]y(t)=sin[x(t)] time-variant?

11.Describe bode plot with example.

Basic factors

Example

Uses

12.Define frequency response of the discrete LTI system. Derive formula to calculate the impulse response of continuous time ideal low pass filter.

Frequency response of a discrete-time LTI system

Impulse response of the CT ideal low-pass filter

13.What is an LTI system? Find the output of the LTI system with impulse response h(t) = e^(−2t)u(t) when the input to the system is x(t) = 1 for |t| < 5; 0 otherwise.

LTI system

Output for h(t)=e−2tu(t)h(t) = e^{-2t}u(t)h(t)=e−2tu(t), x(t)=1x(t) = 1x(t)=1 for ∣t∣<5|t| < 5∣t∣<5

Result

14.Show that ideal low pass filter is not practically realizable.

Step 1: Impulse response

Step 2: Check causality

Step 3: Delay does not help

Step 4: Paley–Wiener criterion

15.What do you mean by impulse response of a LTI system? Derive formula to calculate impulse response of continuous time ideal low pass filter.

Impulse response of an LTI system

Impulse response of the CT ideal low-pass filter

16.State and prove commutative property of continuous time LTI system.

17.Explain invertibility of LTI system and derive the necessary condition for two LTI systems to become inverse of each other.

Condition for LTI systems

Examples

18.Elaborate the physical meaning of convolution integral using suitable example.

19.Derive convolution sum. Prove that an ideal low-pass filter is a non-causal system.

Derivation of the convolution sum

Ideal LPF is non-causal

20.What do you mean by causal and non causal systems? Derive the condition for a continuous time LTI system to be causal.

Causal and non-causal systems

Condition for a CT LTI system to be causal

21.Prove, with necessary derivations, that an ideal low pass filter is not physically realizable. Also derive the expression for step response of ideal low pass filter.

Ideal LPF is not physically realizable

Step response of the ideal LPF

22.Consider a continuous time system with input x(t) and output y(t) related by y(t) = x(t−2) + x(2−t). Is the system Linear Time invariant? Explain.

Linearity

Time invariance

23.Write a short note on LTI system.

24.Write a short note on Bode plot.

25.Derive formula to calculate convolution integral for continuous time LTI system. Find convolution between following continuous time signals x(t) = e^(−t)u(t) and y(t) = e^(−2t)u(t−3)

Derive the convolution integral for continuous time LTI system.

Step 3: Take the limit $\Delta \to 0$

Derive the expression for impulse response of ideal low pass filter (with cutoff frequency ω_c) and discuss.

Evaluate the step response of RC filter.

Describe the frequency response (transfer function) of a continuous time LTI system.

Define linear time invariant system. Explain causality and time variance properties of continuous time system.

Derive the transfer function of a RC low pass filter and plot the magnitude and phase.

Find the output y(t) of LTI system if the impulse response h(t) = e^(−t)u(t−3) and input x(t) = u(t−2).

Step 1: Write both functions in $\tau$

Step 2: Range of $t$

Step 3: Evaluate for $t > 5$

Derive the expression for impulse response of RC filter and plot the magnitude and phase response.

Find the convolution between the signals x(t) = 1 for −1 < t < 1, 0 otherwise and h(t) = 1 for −1 < t < 1, 0 otherwise.

Case 1: $t < -2$

Case 2: $-2 \le t < 0$ (partial overlap from left)

Case 3: $0 \le t < 2$ (partial overlap from right)

Case 4: $t \ge 2$

What are the properties of system? Determine whether the given system is time-variant or not. y(t) = sin[x(t)]

Is $y(t) = \sin[x(t)]$ time-variant?

Describe bode plot with example.

Define frequency response of the discrete LTI system. Derive formula to calculate the impulse response of continuous time ideal low pass filter.

What is an LTI system? Find the output of the LTI system with impulse response h(t) = e^(−2t)u(t) when the input to the system is x(t) = 1 for |t| < 5; 0 otherwise.

Output for $h(t) = e^{-2t}u(t)$ , $x(t) = 1$ for $|t| < 5$

Show that ideal low pass filter is not practically realizable.

What do you mean by impulse response of a LTI system? Derive formula to calculate impulse response of continuous time ideal low pass filter.

State and prove commutative property of continuous time LTI system.

Explain invertibility of LTI system and derive the necessary condition for two LTI systems to become inverse of each other.

Elaborate the physical meaning of convolution integral using suitable example.

Derive convolution sum. Prove that an ideal low-pass filter is a non-causal system.

What do you mean by causal and non causal systems? Derive the condition for a continuous time LTI system to be causal.

Prove, with necessary derivations, that an ideal low pass filter is not physically realizable. Also derive the expression for step response of ideal low pass filter.

Consider a continuous time system with input x(t) and output y(t) related by y(t) = x(t−2) + x(2−t). Is the system Linear Time invariant? Explain.

Write a short note on LTI system.

Write a short note on Bode plot.

Derive formula to calculate convolution integral for continuous time LTI system. Find convolution between following continuous time signals x(t) = e^(−t)u(t) and y(t) = e^(−2t)u(t−3)

Convolution of $x(t)=e^{-t}u(t)$ and $y(t)=e^{-2t}u(t-3)$

Determine whether the given system is linear or not. y(t) = A x(t) + B

Write a short note on causal and non causal systems.

Find output y(t) of LTI system if input x(t) = exp(−at) u(t) and impulse response h(t) = u(t).

Derive the conditions for distortionless transmission for continuous-time LTI system.

What is LTI system? For LTI system, describe the properties: (a) linearity (b) stability (c) time invariance and (d) causality.

Derive Formula to calculate the impulse response of continuous time ideal low pass filter. Is this system practically realizable or not.

Write a short note on invertibility of LTI system.

What is convolution? Obtain the expression for convolution integral.

Find convolution between two signals x(t) = e^(0.5t) for 0 < t < 5, 0 otherwise and h(t) = 1 for 1 < t < 3, 0 otherwise.

Derive the expression for impulse response and step response of ideal low pass filter.

For the LTI system, describe following properties: (a) linearity (b) causality (c) stability (d) time invariance.

The impulse responses of two LTI systems are given by h₁(t) = e^(−t/2)u(t) and h₂(t) = u(t) − u(t−5). Determine the equivalent impulse response if these two systems are connected in cascade. Also sketch the graph of equivalent impulse response.

Let x(t) be the input to an LTI system with unit impulse response h(t), where x(t) = e^(−at)u(t), a > 0 and h(t) = u(t). Verify commutative law of LTI system.

LHS: $y_1(t) = x(t) * h(t)$

RHS: $y_2(t) = h(t) * x(t)$

What is distortionless transmission? Derive the expression for unit step response of ideal low pass filter.

A LTI system has input x(t) = e^(−t)u(t) and impulse response h(t) = e^(t)u(−t). Find output y(t) of the system using Fourier transform of x(t) and h(t).

Write about the following properties of continuous time system: (a) Linearity (b) Causality (c) Memory (d) Stability (e) Time invariance.

Derive the expression for impulse response and step response of first order continuous time system described by the differential equation τ dy(t)/dt + y(t) = x(t).

If the impulse response of continuous time linear time invariant system is h(t) = u(t) − u(t−3) and input to the system is x(t) = u(t+4) − u(t), determine the output y(t) of the system.

Define LTI system. Explain causality and stability property of continuous time LTI system.

Define frequency response of a LTI system. Derive the impulse response of an ideal low pass and high pass filter.