Systems: Definition x[n] y[n] S A system is

Systems: Definition x[n] y[n] S A system is

Systems: Definition x[n] y[n] S A system is a transformation from an input signal . x[n] into an output signal y[n] Example: a filter SIGNAL s[n] NOISE x[n] Filter v[n] y[n] s[n] Systems and Properties: Linearity Linearity: x1[n]

S y1[n] x2 [ n ] S y 2 [ n] x[n] a1 x1[n] a2 x2 [n] S y[n] a1 y1[n] a2 y2 [n] Systems and Properties: Time Invariance if x[n] S y[n] time then

time x[n D] D time y[n D] S D time Systems and Properties: Stability Bounded Output Bounded Input x[n] y[n] S Systems and Properties: Causality x[n]

y[n] S the effect comes after the cause. Examples: y[n] 3 x[n 1] 2 x[n 2] 4 x[n 3] y[n] 3 x[n 1] 2 x[n 2] 4 x[n 3] Causal Non Causal Finite Impulse Response (FIR) Filters x[n] Filter y[n] General response of a Linear Filter is Convolution: N y[n] h[n] * x[n] h[]x[n ] 0 Written more explicitly: y[n] h[0]x[n] h[1]x[n 1] ... h[ N ]x[n N ] Filter Coefficients

Example: Simple Averaging x[n] Filter y[n] 1 y[n] x[n] x[n 1] ... x[n 9] 10 Each sample of the output is the average of the last ten samples of the input. It reduces the effect of noise by averaging. FIR Filter Response to an Exponential Let the input be a complex exponential x[ n] e j0 n Then the output is N y[n] h[]e j0 ( n ) 0 N j0 ) j0 n h[]e e 0

x[n] e j 0 n Filter y[n] H 0 e j0 n Example x[n] e j 0 n Filter y[n] H 0 e j0 n Consider the filter 1 y[n] x[n] x[n 1] ... x[n 9] 10 j 0.1n with input Then x[n] e

1 9 j 0.1 1 1 e j 0.1 10 j1.4137 H 0.1 e 0 . 6392 e j 0.1 10 0 10 1 e and the output y[n] 0.6392e j1.4137 e j 0.1 n Frequency Response of an FIR Filter x[n] e j 0 n Filter y[n] H 0 e j0 n

N H ( ) h[n]e jn n 0 is the Frequency Response of the Filter Significance of the Frequency Response If the input signal is a sum of complex exponentials x[n] X k e jk n k Filter y[n] Yk e jk n k the output is a sum is a sum of complex exponential. Each coefficient is multiplied by the corresponding frequency response: Xk Yk H (k ) X k

Example Consider the Filter x[n] defined as y[n] Filter 1 y[n] x[n] x[n 1] ... x[n 4] 5 Let the input be: x[n] 3 cos(0.1 n 0.2) 2 cos(0.3 n 0.7) Expand in terms of complex exponentials: e 1.0e e x[n] 1.5e j 0.2 e j 0.1 n 1.5e j 0.2 e j 0.1 n j 0.7

j 0.3 n 1.0e j 0.7 j 0.3 n Example (continued) The frequency response of the filter is (use geometric sum) j 5 1 1 1 e H ( ) 1 e j ... e j 4 5 5 1 e j

Then y[n] H 0.1 1.5e j 0.2 e j 0.1 n H 0.1 1.5e j 0.2 e j 0.1 n e H 0.2 1.0e j 0.7 j 0.3 n H 0.2 1.0e j 0.7 e j 0.3 n with H (0.1 ) 0.904e j 0.6283 , H ( 0.1 ) 0.904e j 0.6283 H (0.2 ) 0.647e j1.2566 , H ( 0.1 ) 0.647e j1.2566 Just do the algebra to obtain: y[n] 2.712 cos(0.1 n 0.428) 1.294 cos(0.3 n 1.956) The Discrete Time Fourier Transform (DTFT) Given a signal of infinite duration

x[n] with n define the DTFT and the Inverse DTFT X ( ) DTFT x[n] x[n]e j n n 1 x[n] IDTFT X ( ) 2 Periodic with period 2 jn X ( ) e d X ( 2 ) X ( ) General Frequency Spectrum for a Discrete Time Signal

Since X ( ) is periodic we consider only the frequencies in the interval | X ( ) | F S 2 0 0 (rad ) FS 2 F (Hz ) * X ( ) X ( ) If the signal x[n] is real, then

Example: DTFT of a rectangular pulse Consider a rectangular pulse of length N x[n] 1 0 Then where N1 N1 X ( ) e j n n 0 WN ( ) e n 1 e j N WN ( ) j 1 e j ( N 1) / 2

sin N / 2 sin / 2 Example of DTFT (continued) x[n] 1 0 n N1 DTFT WN ( ) 12 N 10 8 6 4 2 0 -3

-2 -1 0 2 N 1 2 N 2 3 Why this is Important x[n] y[n] Filter

Recall from the DTFT 1 x[n] 2 j n X ( ) e d Then the output 1 jn y[n] X ( ) H

( ) e d 2 Which Implies Y ( ) DTFT y[n] H ( ) X ( ) Summary Linear FIR Filter and Freq. Resp. x[n] y[n] Filter N1 Filter Definition: y[n] h[n] * x[n] h[]x[n ] 0 N1

j n H ( ) h [ n ] e , Frequency Response: n 0 DTFT of output Y ( ) H ( ) X ( ) Frequency Response of the Filter x[n] Filter y[n] Frequency Response:

N1 H ( ) h[n]e jn , n 0 We can plot it as magnitude and phase. Usually the magnitude is in dBs and the phase in radians. Example of Frequency Response Again consider FIR Filter 1 y[n] x[n] x[n 1] ... x[n 9] 10 The impulse response can be represented as a vector of length 10 h 0.1, 0.1 ... 0.1 Then use freqz in matlab freqz(h,1) to obtain the plot of magnitude and phase. Example of Frequency Response (continued) Magnitude (dB) 0

-20 -40 -60 -80 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized Frequency ( rad/sample) 0.9 1 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Normalized Frequency ( rad/sample) 0.9 1 Phase (degrees) 100 0 -100 -200

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