"Solution 1a) Given:? (? ) = 0.6? (? )- 0.1? (? - 2)- 0.2? (? - 3)? in(2* pi/12* n) + cos(2* pi/6* n) + cos(2* pi/24* n) for n = 0 to 24? (? ) =cos(2* pi/6* n)for n = 25 to 47 0 otherwise Now we can write it as convolution form? ? =? ? *h(? )Matlab Programclc;clear all;close all;% define nn=0:60;% define x(n)for n=0:60if(n>0 && n< 25)x(n+1)=sin(2*pi/12*n)+cos(2*pi/6*n)+cos(2*pi/24*n);endif(n>24 && n< 48)x(n+1)=cos(2*pi/6*n);endif(n>47)x(n+1)=0;endend % define y(n)y(1)=0.6*x(1);y(2)=0.6*x(2)-0.1*x(1);for n=3:60y(n+1)=0.6.*x(n+1)-0.1.*x(n-1)-0.2.*x(n-2);end%% plotsfigure;n=0:60;plot(n,x)xlabel('n--------------------->')ylabel('x(n)------------------>')title('x(n)')grid on;saveas(gcf,'x(n).jpg','jpg')figure;n=0:60;plot(n,y)xlabel('n--------------------->')ylabel('y(n)------------------>')title('y(n)')grid on;saveas(gcf,'y(n).jpg','jpg')Plots b) Poles are at z=0,Zeroes are at 0, -1+2j, -1-2jMatlab program%% Part b%TF z transformsymsz;TF=0.6-0.1*z^-2-0.2*z^-3;F2 = collect(TF)[num,den] = numden(F2);Ts = 0.1; % Sampling periodH = tf(sym2poly(num),sym2poly(den),Ts)figure;pzmap(H) Solution 2Causal SystemThe system is as we can clearly see that it is dependent on the future values of n. ? ? =? ? ? {? ? - 3 ,? ? - 2 ,? ? - 1 ,? ? ,? ? + 1 ,? ? + 2 ,? ? + 3As we discussed the system dependency on future and past it is a non causal systemLinear SystemUsing homogeneity and additivity it can be depicted that system can satisfy these both partsand hence it is linear ? ? =? ? ? {? ? - 3 ,? ? - 2 ,? ? - 1 ,? ? ,? ? + 1 ,? ? + 2 ,? ? + 3Time Invariant System The system is time variant as we can clearly see that it is dependent on the future values of n.? ? =? ? ? {? ? - 3,? ? - 2,? ? - 1,? ?,? ? + 1,? ? + 2,? ? + 3So clearly we need to find ? ? + 1,? ? + 2,? ? + 3 for this and which future values hence systemis not time invariant is.Solution 3a) A perfect frequency selective filter passes complex exponential information signal, for a givenarrangement of frequencies and totally rejects the others. The perfect ideal filter has a recurrencereaction that is real and positive, as it were, has a zero phase qualities. A linear phase qualities presentsa period shift and this causes no distortion in shape of the flag in the pass-band. Because the Fouriertransform of a steady impulse reaction is continuous function of ?, cannot get a stable ideal filter. The ideal perfect filter with limited bands of zero reaction can't be executed with a causal channel. We should configuration filters that surmised the sought frequency reaction H (?). The perfect ideald recurrence reaction H (?) is spread out by the recurrence reaction W(?) of the window work.d The system function –? ? ? ? ? ? ? = ? ? -1 ? ?1-? ? ? =1 ? ? ? ? A pole at s=s in the s-plane is transformed to a pole at z=? in the z-plane. If the comtinuous timek filter is stable, ? ? ? < 0? ? ? ? ? Then, the magnitude of ? will be less than 1,So the pole will be in the unit circle. Thus the causal discrete time filter is stable.LTI framework is alluded to as a frequency selective filter, which posses thereceived signal to wipe outthehigh-recurrence segments presented by the channel obstruction and commotion from the low- recurrence segments constituting the data bearing sign. The scope of frequencies dispensed with fromthe CT flag connected at the information of the channel is alluded to as the stop band of the channel, while the scope of frequencies that is left moderately unaffected by the channel constitute the passband of the channel. We are especially intrigued in the recurrence particular channels that are classified in four distinctclassifications (lowpass, highpass, bandpass, and bandstop).As is valid for any perfect framework, the ideal frequency selective filters are most certainly notphysically feasible for an assortment of reasons. 1. From the recurrence qualities of the ideal filters, we take note of that the gain A of the filter issteady in the pass band, while the gain in the stop band is entirely zero. 2. Next issue with the transfer function H(w), indicated for perfect filters is the sharp transitionbetween the pass and stop band with the end goal that there is a intermittence in H(w) at w=wc. By and by, we can't actualize filter with steady gains in the pass and stop band. Likewise, suddentransition can't be outlined. This can be the steady gain and the sharp transition in the perfect low- pass and high-pass channels prompt to non-causal impulse reactions which are of infinite length.Obviously, such LTI frameworks can't be actualized in the physical world.b) From the question we can notice that:X(Z) has two zeroes at the origin which means that it has two zeroes at the same point. So the2 numerator will be ?? 1 ? 3 Also we can notice that X(Z) has two poles? and ? and one of these poles = ?1 2 2 ? 1 1 1 1.732 ? 3 ? = * +? 2 2 2 2 So 2 2 ? ? X(Z)== ? 1 ? ? -? (? -? ) 1 2 3 ? -? (? -? ) 1 2 8 Given that X(1) = 3 If we apply to the equation it will be2 1 1 8 X(1) = == ? ? 1 1 ? ? 3 33 1-? (1-? ) 1-? (1-? ) 1 1 2 2Solving this? ? 1 1 8 ? ? 3 3 1-? - ? +?? = 1 1 2 2 3 ? ? 1 8 1 ? ? 3 3 ?? - ? =+ ? – 11 1 2 3 2 ? ? 1 8 1 ? ? 3 3 ? (? - 1) =+ ? – 11 3 2 2 ? ? 8 1 ? 1 ? 5 3 3+ ? - 1 ? +3 2 2 3 ? == ? ? 1 1 ? 1 ? 3 3? - 1? - 1 2 2 ? 1 5 ? 3 ? +2 3 ? =1 ? 1 ? 3? - 1 2 1 1 1.732 5 * +? +2 2 2 3 ? =1 1 1 1.732 * +? - 1 2 2 2 23 1.732 +? 12 4 ? =1 3 1.732 - +? 4 4 23 1.732 +? 1.4375- 0.1875 +? (1.15) 12 4 ? = =1 3 1.732 -0.75 - +? 4 4 23 1.732 +? 12 4 ? = =-(1.67 +? 1.533)1 3 1.732 - +? 4 4 So finally2 2 ? ? X(Z)==? -?(? -? ) ? -?(? -) 1 2 1 2 ? X(Z)= 1 1 1.732? +1.67+? 1.53? - * +? 2 2 2Roc –Answer 4a) Kaiser window -2 2? ? ? =? ? ? 1- - 1; 0=? =? - 10 ? - 1Given that,Low-pass filter with –Order N=63Pass-band w =0.3?p Stop-band w =0.32?s ?? =?s –?p=0.02? ? =-20 log ? ? ?? ? The number of cycles of ripple in the pass band is equal to the order of the filter. ? - 8 ? ? =2.285 ??? = 63* 2.285* 0.02* 3.14 + 8 = 17.04? ? ?? ? ? = 24.63 ? ?? If ds = 0.01Then,? ?? ? ? =-20 log ? ? = 40? ??b) ? =?? ? If ? =? = 6% and ? ? Then,? ?? ? ? =-20 log 0.06 = 24.4 ? ??"