Frequency-Domain C/S of LTI Systems x(n) y(n) LTI I LTI: Linear Time-Invariant system I h(n), the impulse response of an LTI systems describes the time domain c/s. I H(!), the frequency response describes the frequency-domain c/s. F I h(n) −−! H(!) I study: system response to excitation signals that are a weighted linear combination of sinusoids or complex exponentials. 1/ 25 Frequency-Domain C/S of LTI Systems I Recall the response of an LTI system to input signal x(n) 1 X y(n) = h(n)x(n − k) (1) k=−∞ 2/ 25 Frequency-Domain C/S of LTI Systems I Excite the system with a complex exponential, i.e. let x(n) = Aej!n; −∞ < n < 1; −π < ! < π 1 X h i y(n) = h(k) Aej!(n−k) k=−∞ " 1 # X (2) = A h(k)e−j!k ej!n k=−∞ = Aej!nH(!) where, 1 X H(!) = h(k)e−j!k (3) k=−∞ 3/ 25 Frequency-Domain C/S of LTI Systems Observations I y(n) is in the form of a complex exponential with same frequency as input, multiplied by a factor. I The complex exponential signal x(n) is called an eigenfunction of the system. I H(!) evaluated at the frequency of the input is the corresponding eigenvalue of the system. 4/ 25 Frequency-Domain C/S of LTI Systems Observations, cont. I As H(!) is a Fourier transform, it is periodic with period 2π I H(!) is a complex-valued function, can be expressed in polar form H(!) = jH(!)jeφω I h(k) is related to H(!) through π 1 Z h(k) = H(!)ej!k d! 2π −π 5/ 25 Frequency-Domain C/S of LTI Systems Real-valued impulse response An LTI system with a real-valued impulse response, exhibits symmetry properties as derived in section 4.4.1 I jH(!)j ) even function of ! I φ(!) ) odd function of !. I Consequently, it is enough to know jH(!)j and φ(!) for −π ≤ ! ≤ π, 6/ 25 Example I Determine the output sequence of the system with impulse response 1 h(n) = ( )nu(n) 2 when the input signal is x(n) = Aejπn=2; −∞ < n < 1 7/ 25 Example, cont. I Solution: evaluate H(!) 1 X 1 H(!) = h(n)e−j!n = 1 −j! n=−∞ 1 − 2 e π 1 2 o H(! = ) = = p e−j26:6 2 1 1 + j 2 5 I Thus, output is 2 o y(n) = A p e−j26:6 ejπn=2 5 2 o = p Aej(πn=2−26:6 ); −∞ < n < 1 5 I system effect on the input: amplitude scale and phase shift I change frequency, we change amount of scale change and phase shift. 8/ 25 Example: Moving Average Filter I Determine the magnitude and phase of H(!) for the three-point moving average (MA) system. 1 y(n) = [x(n + 1) + x(n) + x(n − 1)] 3 and plot these functions for 0 ≤ ! ≤ π 9/ 25 Example: Moving Average Filter, cont. I Solution: 1 1 1 h(n) = f ; ; g 3 3 3 consequently, X 1 1 H(!) = h(k)e−j!k = (ej! + 1 + e−j!) = (1 + 2 cos !) 3 3 n Hence 1 jH(!)j = j1 + cos !j 3 8 2π <> 0; 0 ≤ ! ≤ φ(!) = 3 2π :> π; ≤ ! < π 3 10/ 25 11/ 25 Example: system response to sinusoids I Determine the response of the system in Example 5.1.1, for the input signal π x(n) = 10 − 5 sin n + 20 cos πn; −∞ < n < 1 2 I Recall, the frequency response of the system is 1 H(!) = 1 −j! 1 − 2 e 12/ 25 Example, cont. I Solution I Idea Recognise the frequency of each part of the input signal, and find corresponding system response. I First term 10, fixed signal ) ! = 0 1 H(0) = 1 = 2 1 − 2 π I 5 sin 2 n has a frequency ! = π=2, thus π 2 o H( ) = p e−j26:6 2 5 I 20 cos πn has a frequency ! = π 2 H(π) = 3 13/ 25 Frequency Domain c/s of LTI systems The General Case I Most general case: input to the system is an arbitrary linear combination of sinusoids of the form L X x(n) = Ai cos(!i n + φi ); −∞ < n < 1 i=1 Where fAi g and fφi g amplitude and phase of corresponding sinusoidal component i . I System response will be of the form: L X y(n) = Ai jH(!i )j cos [!i n + φi + Θ(!i )] i=1 Where jH(!i )j and Θ(!i ) are the magnitude and phase imparted by the system to the individual frequency components of the input signal. 14/ 25 Steady-State and Transient Response to Sinusoidal Input Signals I If excitation signal (exponential or sinusoidal) applied at some finite time instant,e.g. n = 0 response = steady-state + transient I Example: let y(n) = ay(n − 1) + x(n) system response to any x(n) applied at n = 0 n X y(n) = an+1y(−1)+ ak x(n−k); n ≥ 0; y(−1) initial condition k=0 I Let x(n) be a complex exponential x(n) = Aej!n; n ≥ 0 15/ 25 Steady-State and Transient Response to Sinusoidal Input Signals, cont. I We get, n X y(n) = an+1y(−1) + A ak ej!(n−k) k=0 " n # X = an+1y(−1) + A (ae−j!)k ej!n k=0 1 − an+1e−j!(n+1) = an+1y(−1) + A ej!n; n ≥ 0 1 − ae−j! Aan+1e−j!(n+1) A = an+1y(−1) − ej!n + ej!n 1 − ae−j! 1 − ae−j! 16/ 25 Steady-State and Transient Response to Sinusoidal Input Signals, cont. I BIBO stable, if jaj < 1 n+1 I Since jaj < 1, the terms containing a ! 0, as n ! 1. I Steady-state response: A j!n yss (n) = lim y(n) = e n!1 1 − ae−j! = AH(!)ej!n I Transient response of the system: Aan+1e−j!(n+1) y (n) = an+1y(−1) − ej!n; n ≥ 0 tr 1 − ae−j! 17/ 25 Steady-State Response to Periodic Input Signals I Let the input signal x(n) to a stable LTI be periodic with fundamental period N. I Periodic ! −∞ < n < 1 ! total response of the system is the steady-state response. I Using the Fourier series representation of a periodic signal N−1 X j2πkn=N x(n) = ck e ; k = 0; 0; :::; N − 1 k=0 18/ 25 Steady-State Response to Periodic Input Signals, cont. I Evaluating the system response for each complex exponential j2πkn=N xk (n) = ck e ; k = 0; 1; :::; N − 1 2πk y (n) = c H ej2πkn=N ; k = 0; 1; :::; N − 1 k k N where 2πk H( ) = H(!)j ; k = 0; 1; :::; N − 1 N !=2πkn=N 19/ 25 Steady-State Response to Periodic Input Signals, cont. I Superposition principle for linear systems: N−1 X 2πk y(n) = c H ej2πkn=N ; −∞ < n < 1 k N k=0 I LTI system response to a periodic input signal is also periodic with the same period N, with coefficients related by 2πk d = c H ; k = 0; 1; :::; N − 1 k k N 20/ 25 Response to Aperiodic Input Signals I Let fx(n)g ne the aperiodic input sequence, fy(n)g output sequence, and fh(n)g unit sample response. I By Convolution theorem Y (!) = H(!)X (!) In polar form, magnitude and phase of the output signal: jY (!)j = jH(!)jjX (!)j \Y (!) = \H(!) + \X (!) I Output of an LTI system can NOT contain frequency components that are not contained in the input signal. 21/ 25 Response to Aperiodic Input Signals, cont. I Energy density spectra of input and output jY (!)j2 = jH(!)j2jX (!)j2 2 Syy (!) = jH(!)j Sxx (!) I Energy of the output signal π 1 Z E = S (!)d! y 2π yy −π π 1 Z = jH(!)j2S (!)d! 2π xx −π 22/ 25 Example I A linear time-invariant system is characterized by its impulse response 12 h(n) = u(n) 2 I Determine the spectrum and energy density spectrum of the output signal when the system is excited by the signal 12 x(n) = u(n) 4 23/ 25 Example: Solution I Frequency response of the system 1 2 X 1 H(!) = e−j!n 2 n=0 1 = 1 −j! 1 − 2 e I Similiarly, Fourier transform of the input sequence 1 X (!) = 1 −j! 1 − 4 e I Spectrum of the output signal Y (!) = H(!)X (!) 1 = 1 −j! 1 −j! (1 − 2 e )(1 − 4 e ) 24/ 25 Example, cont. I energy density spectrum 2 2 2 Sxx (!) = jY (!)j = jH(!)j jX (!)j 1 = 5 17 1 ( 4 − cos !)( 16 − 2 cos !) 25/ 25.