--- title: "DT transfer function; drawing a Bode plot" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:dt-transfer-function-bode" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-transfer-function-bode site: BCT Engineering --- # DT transfer function; drawing a Bode plot Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2078 Poush, 2+3 marks - Video: [How to Draw Bode Plots Given Transfer Function (Part 1)](https://www.youtube.com/watch?v=6BsCTrAUstI) — Raiya Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-transfer-function-bode - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question Define transfer function and impulse response in discrete-time. Briefly explain the procedure of drawing a bode plot. ## Short answer Impulse response h[n] is the output for δ[n] input. Transfer function H(z)=Y(z)/X(z) equals Z-transform of h[n]. Bode plot procedure: factor H(s) into standard form, identify corner frequencies, draw asymptotic magnitude (dB) and phase (degrees) plots by summing contributions, then correct. ## Answer #### Impulse response (discrete time) The **impulse response** $h[n]$ of a discrete-time LTI system is the output when the input is the unit impulse $\delta[n]$ and the system is initially at rest: $$ \delta[n] \rightarrow h[n] $$ Because any input can be written as $x[n]=\sum_k x[k]\delta[n-k]$, the output of an LTI system is fully fixed by $h[n]$ through the convolution sum $y[n]=x[n]*h[n]$. #### Transfer function (discrete time) The **transfer function** is the ratio of the transform of the output to the transform of the input (zero initial conditions). It equals the transform of $h[n]$: $$ H(z)=\frac{Y(z)}{X(z)}=\sum_{n=-\infty}^{\infty}h[n]z^{-n}, \qquad H(e^{j\omega})=\frac{Y(e^{j\omega})}{X(e^{j\omega})}=\sum_{n}h[n]e^{-j\omega n} $$ $H(e^{j\omega})$ (the frequency response) is $H(z)$ evaluated on the unit circle $z=e^{j\omega}$. Example: for $y[n]=0.5y[n-1]+x[n]$, $H(e^{j\omega})=\frac{1}{1-0.5e^{-j\omega}}$ and $h[n]=(0.5)^n u[n]$. #### Procedure for drawing a Bode plot A Bode plot is a pair of graphs: magnitude in dB, $20\log_{10}|H(j\omega)|$, and phase $\angle H(j\omega)$, both against $\log_{10}\omega$. 1. Write $H(j\omega)$ in **standard (time-constant) form**, e.g. $H(j\omega)=\frac{K(1+j\omega/z_1)}{j\omega(1+j\omega/p_1)}$. 2. Identify the factors: constant $K$, poles/zeros at the origin, simple real poles/zeros, and quadratic factors. Note each **corner frequency** ($\omega=z_1, p_1,\dots$). 3. Draw the asymptote of each factor: - constant: flat line at $20\log K$ dB, phase 0° (or 180° if $K<0$); - pole at origin: line of $-20$ dB/decade through 0 dB at $\omega=1$, phase $-90^\circ$; - simple pole: 0 dB up to the corner, then $-20$ dB/decade; phase goes from 0° to $-90^\circ$ (−45° at the corner); - simple zero: same but $+20$ dB/decade and $+90^\circ$; - quadratic pole: $-40$ dB/decade after $\omega_n$, phase 0° to −180°. 4. **Add** the magnitude asymptotes (dB adds because log of a product is a sum), starting from the lowest frequency; the slope changes by ±20 dB/dec at each corner. 5. **Add** the phase curves of all factors at several frequencies. 6. Correct the asymptotic plot if needed (about 3 dB at a simple corner; resonance peak for low damping). For a discrete-time system the same idea is used with $H(e^{j\omega})$ plotted for $0\le\omega\le\pi$ (it is periodic in $2\pi$), usually on a linear frequency axis. ```text dB |------\ magnitude: flat, then | \ -20 dB/dec rolls off after corner | \ +---------+--------> log w w_c ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The impulse response h[n] of a DT LTI system is: A. The output for a unit step input B. The output for delta[n] input with the system initially at rest C. The ratio Y(z)/X(z) in time domain after a step D. The input that gives zero output Answer: B. h[n] is defined as the response to the unit impulse delta[n]. (Key point: Impulse response) 2. The transfer function H(z) is: A. Y(z)/X(z) with zero initial conditions, equal to the z-transform of h[n] B. X(z)/Y(z) C. The z-transform of the input D. h[n] evaluated at z = 1 Answer: A. H(z) = Y/X = sum h[n]z^(-n); on the unit circle it becomes the frequency response. (Key point: Transfer function) 3. H(e^jw) is obtained from H(z) by: A. Setting z = 0 B. Evaluating on the unit circle z = e^(jw) C. Setting z = 1 only D. Taking the inverse transform Answer: B. The frequency response is H(z) on the unit circle. (Key point: Frequency response link) 4. On a magnitude Bode plot, a simple pole at its corner frequency causes the slope to change by: A. +20 dB/decade B. -20 dB/decade C. -40 dB/decade D. -6 dB/decade at the origin Answer: B. A simple pole gives 0 dB up to the corner and then -20 dB/decade, with phase -45 degrees at the corner. 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