--- title: "Transfer function of y[n] = 0.4y[n−1] + x[n]" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:difference-equation-0-4" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-0-4 site: BCT Engineering --- # Transfer function of y[n] = 0.4y[n−1] + x[n] Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2079 Jestha, 7 marks - Video: [Given the IIR filter: Determine the transfer function, nonzero coefficients, and impulse response](https://www.youtube.com/watch?v=JXGAz_1Dgas) — Engg-Course-Made-Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-0-4 - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question For a system characterized by linear constant coefficient difference equation y[n] = 0.4y[n−1] + x[n], find and plot the transfer function and impulse response of the system. ## Short answer Transfer function: H(e^jω) = 1/(1 − 0.4e^{−jω}), a low-pass filter. Impulse response: h[n] = (0.4)^n u[n], causal and stable. ## Answer **Given:** $y[n] = 0.4\,y[n-1] + x[n]$, i.e. $y[n] - 0.4\,y[n-1] = x[n]$, initially at rest. #### Transfer function (frequency response) Take the DTFT of both sides ($y[n-1] \leftrightarrow e^{-j\omega}Y(e^{j\omega})$): $$ \begin{aligned} Y(e^{j\omega})\left(1 - 0.4e^{-j\omega}\right) &= X(e^{j\omega}) \\ H(e^{j\omega}) &= \frac{Y(e^{j\omega})}{X(e^{j\omega})} = \frac{1}{1 - 0.4e^{-j\omega}} \end{aligned} $$ (In the z-domain, $H(z) = \frac{1}{1 - 0.4z^{-1}} = \frac{z}{z - 0.4}$, a pole at $z = 0.4$ inside the unit circle.) With $e^{-j\omega} = \cos\omega - j\sin\omega$: $$ |H(e^{j\omega})| = \frac{1}{\sqrt{(1 - 0.4\cos\omega)^2 + (0.4\sin\omega)^2}} = \frac{1}{\sqrt{1.16 - 0.8\cos\omega}} $$ $$ \theta(\omega) = -\tan^{-1}\frac{0.4\sin\omega}{1 - 0.4\cos\omega} $$ | $\omega$ | 0 | $\pi/4$ | $\pi/2$ | $3\pi/4$ | $\pi$ | |---|---|---|---|---|---| | $\lvert H\rvert$ | 1.667 | 1.297 | 0.928 | 0.761 | 0.714 | | $\theta$ | 0° | −21.5° | −21.8° | −12.4° | 0° | ```text |H| 1.67| * | .' '. 1.30| * * 0.93| * * 0.71| * * +--+-------+-------+--> w -pi 0 pi ``` The magnitude is even and periodic ($2\pi$), maximum $1/0.6 = 1.667$ at $\omega = 0$ and minimum $1/1.4 = 0.714$ at $\omega = \pi$: a **low pass** system. #### Impulse response Using $a^n u[n] \leftrightarrow \frac{1}{1 - ae^{-j\omega}}$ with $a = 0.4$: $$ h[n] = (0.4)^n\,u[n] $$ | $n$ | <0 | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---|---| | $h[n]$ | 0 | 1 | 0.4 | 0.16 | 0.064 | 0.0256 | (Recursion check: $h[0] = 1$, $h[1] = 0.4h[0] = 0.4$, $h[2] = 0.16$.) ```text h[n] 1.0 | o | | 0.4 | | o 0.16 | | | o | | | | o . 0 +--+--+--+--+--+--> n 0 1 2 3 4 ``` The system is causal, stable ($\sum h[n] = 1/0.6 = 1.667$) and IIR. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Taking the DTFT of y[n] = 0.4y[n-1] + x[n] gives H(e^jw) = A. 1/(1 + 0.4e^(-jw)) B. 1/(1 - 0.4e^(-jw)) C. 1 - 0.4e^(-jw) D. 0.4/(1 - e^(-jw)) Answer: B. y[n-1] becomes e^(-jw)Y, so Y(1 - 0.4e^(-jw)) = X. (Key point: Transfer function) 2. The impulse response of the system is: A. h[n] = (0.4)^n u[n] B. h[n] = (-0.4)^n u[n] C. h[n] = 0.4 delta[n] D. h[n] = (0.4)^n u[-n] Answer: A. a^n u[n] has DTFT 1/(1 - a e^(-jw)); with a = 0.4 and recursion h[0]=1, h[1]=0.4. (Key point: Impulse response) 3. |H| = 1.667 at w = 0 and 0.714 at w = pi. The system is: A. High pass B. Low pass C. Band pass D. All pass Answer: B. Gain is largest at low frequency and smallest at high frequency. (Key point: Filter type) 4. Why is the system stable? A. The pole z = 0.4 is inside the unit circle (sum of h is finite) B. It is FIR C. It has a zero at the origin D. The input is bounded Answer: A. H(z) = z/(z - 0.4) has a pole at 0.4 with |0.4| < 1, so sum of |h[n]| = 1.667 is finite. 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