--- title: "Four properties of DTFS" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-four-properties" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-four-properties site: BCT Engineering --- # Four properties of DTFS Signals and Systems · Chapter 2: Fourier Series - Asked in: 2078 Poush, 6 marks - Video: [Linearity | Properties of DTFS | EnggClasses](https://www.youtube.com/watch?v=GpK3B8KK7Ic) — EnggClasses - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-four-properties - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question State and explain four properties of discrete time Fourier series. ## Short answer Linearity: A x[n]+B y[n] ↔ A a_k+B b_k. Time shift: x[n-n0] ↔ a_k e^{-jkω0 n0}. Frequency shift: e^{jMω0 n}x[n] ↔ a_{k-M}. Parseval: (1/N)∑|x[n]|^2 = ∑|a_k|^2. ## 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: $a_{k+N} = a_k$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If x[n] has coefficients a_k and y[n] has b_k, the coefficients of A x[n] + B y[n] are: A. A a_k b_k B. A a_k + B b_k C. (A + B)(a_k + b_k) D. A b_k + B a_k Answer: B. The analysis sum is linear, so the coefficients of a weighted sum are the same weighted sum of coefficients. (Key point: Linearity) 2. What does a time shift x[n - n0] do to the DTFS coefficients? A. Shifts a_k to a_{k-n0} B. Changes |a_k| and leaves the phase unchanged C. Multiplies a_k by e^{-jk(2 pi/N)n0}, changing only the phase D. Multiplies a_k by n0 Answer: C. A shift multiplies a_k by a linear-phase term, so |a_k| is unchanged and only the phase changes. (Key point: Time shifting) 3. Multiplying x[n] by e^{jM(2 pi/N)n} gives coefficients: A. a_{k+M} B. a_k e^{jM} C. M a_k D. a_{k-M} Answer: D. Multiplying by the Mth harmonic exponential shifts the coefficient sequence by M places, giving a_{k-M}. (Key point: Frequency shifting) 4. Parseval's relation for the DTFS states that the average power in one period (1/N) sum |x[n]|^2 equals: A. |a_0|^2 only B. sum of a_k over one period C. N times the sum of |a_k|^2 D. sum of |a_k|^2 over the N harmonics Answer: D. Average power equals the sum of the powers |a_k|^2 of the N harmonic components. 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