MIT OpenCourseWare http://ocw.mit.edu MAS.160 / MAS.510 / MAS.511 Signals, Systems and Information for Media Technology Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � Frequency response FIR M yn[]= kxn� k]b[ k= 0 M hk[]=bk�[n � k]k=0 M H ˆ= [k]ej ˆk()hk=0 M H ()ˆ=bj ˆk kek=0 Easy to go from difference equation to frequency response because h[n] finite length and h[n] = [b0, b1,…]. IIR N M yn =a yn� l +b xn� k []l[ ] k[ ]l=1 k=0 hk[]��bk[n � k]k=0 Argh! x H ()ˆ=�hj�ˆk�[k]ek=0 Tough to go from diff.eqn. to freq. response because h[n] infinite length, h[n]=f(al,bk) is complicated, and �ˆmay be unbounded.H ()� � � � �  � ( ) temporal space - n complex frequency space- z N M yn =a yn� l +b xn� k bkz� k �(z � z)[]l[ ] k[ ] MM l=1 k=0 ziHz= k=0 =i=0()N N hk [] bk[n � k] z-transform � k 1� z � z a z k pi k=0 i=0 k=1 Argh!Fourier transform @#! road block j�ˆk H �ˆ= h[k]e ()j� ()k=0 z=eHurray! The frequency response is H � ()He Hz ()= = j� frequency space - � H(z) evaluated on unit circle Benefits of z-plane and z-transforms: 1. Get around road block by using z-plane and z-transforms. Compute system function from diff.eq. coefficients, then evaluate on the unit circle to find the frequency response. 2. z-plane (pole/zeros) will tell us if system stable and frequency response exists. 3. By using z-transforms, solution to diff.eq goes from solving convolution in n-space to solving algebraic equations in z-domain (easier). And lots more…!� Infinite signals x[n] = anu[n] � X(z) =�akz�k x[n]=0 n<0 k= 0 2 3 right sided =1+ az�1 +(az�1)+(az�1)K geometric series Xz=1+ az�1 + az�12 + az�13 K()( ) ( )az�1Xz= � az�1 �(az�12 �(az�13 �(az�14 K-()) ) ) (1� az�1)Xz= ()1 1 �1 <1 region of convergenceXz= az()1� az�1 z > aorInfinite signals �1 x[n] = �anu[�n �1]� X(z) = ��akz�k x[n] = 0 n � 0 k=�� = � 1 z � 1 z2 � 1 z3 K() ()left sided a a a geometric series 12 13Xz()= � 1 z � z �()K()za a a -az�1Xz=1+1 z +1 z2 +1 z3 +1 z4 K()() () ()a a a a (1� az�1)Xz=1()1 1 z <1 region of convergenceXz= a()1� az�1 z < aorInfinite series: x[n] =nu[n] � ()= 1 aXzregion of convergence z > a1� az�1 x[n]=0 n<0 right sided x[n] = �anu[�n �1] � Xz= 1 z < a()1� az�1 x[n] = 0 n � 0 left sided Finite series: N �1 x[n] = an (u[n � M] � u[n � N])� X(z) =akz�k k= M M N(az�1)�(az�1) Xz= ()1� az�1 all z region of convergence lim X(z) = N � M z�a1 Equivalent ways to represent the system unit delay++ b1 a2 1N M x[n] b0yn =alyn� l +bkxn� k �[] [ ] [ ]l=1 k=0 difference equation y[n]block diagramc inspection x[n]=�[n] c MM 3 z bkz�k �(z � z)h[n] = y[n] x[n ]=�[ n ] ()= k=0 N =iN =0� Hzzi impulse response 1�akz�k �(z � z) k=1 i=0 pi sequence system function pole-zero polynomial locations 54 j� z = e c 6 All poles must be inside unit circle for ()H �H ()�=()= HzHej�()z=ej� to converge and the system frequency response to be stable. (causal system) (FIR filter always stable)1 Equivalent ways to represent the system unit delay++ b1 a2 1N M x[n] b0yn =alyn� l +bkxn� k �[] [ ] [ ]l=1 k=0 difference equation y[n]block diagramc inspection x[n]=�[n] c MM 3 z bkz�k �(z � z)zih[n] = y[n] x[n ]=�[ n ] � Hz= k=0 N =iN =0()impulse response 1�akz�k �(z � z)k=1 i=0 pi sequence 4 system function pole-zero polynomial locations 5 j� z = e c 6 The region of convergence must Contain the unit circle for ()H �H �= Hej�= Hz()()()z=ej� to converge and the system frequency response to be stable. (general) (FIR filter always stable)Ex. num=0 denom=0 5 4 3 |H(z)| 2 1 0 2 z2 + z +1()13 �13 �23zHz = +13 z +1 z = 2 y[n] =()nHzzHz = 0 y[n] = 0 z + z +1 = 0 zeros()2 ± j 2� 1 3z =(�1± j 3)= e roots of numerator 2 () 2 Hz = � y[n] = � z = 0 poles z = 0, 0 roots of denominator *FIR L point summer/averager only has zeros on unit circle FIR filters only have zeros on unit circle, and poles are either at 0 or �. #poles=#zeros “extra” zero/poles are at z=�. -2 -2 -1 0 1 2 -1 0 1 2 Im(z) Re(z)25-pt averager lowpass FIR filter *poles all at zero (or �) *zeros evenly distributed on unit circle *missing zero at DC (lowpass) 24-pt bandpass FIR filter b=fir1(24, [.45 .65],'bandpass'); *poles all at zero (or �) *zeros not necessarily on unit circle * Only pole locations affect stabilityEx. Hz =1 +1 z�1 +1 z�2 = z2 + z +1()3 3 33z2 y[n] = Hzzn ()num=0 Hz = 0 y[n] = 0 z2 + z +1 = 0 zeros()± j 2� 31z =2 (�1± j 3)= e roots of numerator 2denom=0 Hz = � y[n] = �()z = 0 poles 5 4 3 |H(z)| 2 1 0 2 -1 0 1 2 -1 0 1 2 Im(z) Re(z) z = 0, 0 roots of denominator FIR filters have poles at either at 0 or �. #poles=#zeros “extra” zero/poles are at z=¥. -2 -2� system response |H(z)| pz plot 1 5 0.8 4 0.6 0.4 3 0.2|H(z)| 2 0 1 -0.2 2 -0.4 0 Imaginary part2 2 TextEnd 2 -0.6 1 2 -0.8 1Im(z) 0 -1 -10 Re(z) -1 -1 -0.5 0 0.5 1 -2 -2 Real part frequency response H �= Hej�= Hz()()()j�z=eThe frequency response is H(z) evaluated on unit circle -2 -1 0 1 2 3 1 0.9 0.8 0.7 |H(ej�)|0.6 0.5 0.4 0.3 0.2 0.1 0 -3Solving impulse response y[n] = x[n] � y[n � 2] x[n] =�[n] z-transform iteration y[0] = x[0]� y[�2] =1� 0 =1 Y (z) = X (z) � z�2Y (z) y[1] = x[1] � y[�1] = 0 � 0 = 0 (1+ z�2 )Y (z) = X (z) y[2] = x[2]� y[0] = 0 �1 = �1 y[3] = x[3]� y[1] = 0 � 0 = 0 Y (z) 1 system y[4] = x[4]� y[2] = 0 � (�1) = 1 H (z) = X (z) =(1+ z�2 ) function M 1Y (z) = H (z)X (z) = X (z)(1+ z�2 ) Remember: M 1 z2 2 =bkz�kz � cos( ˆ)zY (z) = Hz = X(z) = k= 0 X (z) =�22()Y(z)1� N akz�kz2 � 2 cos( ˆ)z +1 (1+ z) z +1 k=1 c � Inverse z-transform (lookup) x[n] =�n[]cos(ˆ)un� 2�� � 2� y[n] = h[n] = cos� nu[n] ˆ= = � 4 �2 4 y[n] = {1, 0, �1,0,1,K}Solve difference equation � 2�� y[n] = x[n] � y[n � 2] h[n] = cos � nu[n] impulse response � 4 � x[n] = u[n] step input iteration y[0] = x[0]� y[�2] = 1� 0 = 1 y[1] = x[1] � y[�1] = 1� …