DeCarlo Ch20 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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SOLUTIONS PROBLEMS CHAPTER 20 U SEFUL MATLAB M - FILES FOR USE IN THE SOLUTION TO PROBLEMS IN THIS CHAPTER . Program 1: converts y-parameters to t-parameters % convert y parameters to t parameters function [t,t11,t12,t21,t22] = ytot(y) y11=y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2); deltay= y11*y22-y12*y21; t11=-y22/y21; t12 = -1/y21; t21= -deltay/y21; t22= -y11/y21; t= [ t11 t12; t21 t22]; Program 2: Computes Zin, Zout, and gains using two port h-parameters. »% two-port analysis in terms of h-parameters »function [zin, zout] =twoport(h, zL, zs) »['twoport analysis using h-parameters'] »h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2); »zin = h11 - h12*h21/(h22+ 1/zL) »yout= h22 - h12*h21/(h11+zs); »zout= 1/yout »v1tovs= zin/(zin+zs) »v2tov1= -h21/(zin*(h22+1/zL)) »v2tovs= v1tovs*v2tov1 Program 3: Computes Zin, Zout, and gains using two port y-parameters. % two-port analysis in terms of y-parameters function [zin, zout] =twoporty(y, zL, zs) ['twoport analysis using y-parameters'] y11= y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2); yin = y11 - y12*y21/(y22+ 1/zL) zin= 1/yin yout= y22 - y12*y21/(y11+1/zs) zout= 1/yout v1tovs= zin/(zin+zs) v2tov1= -y21/(y22+1/zL) v2tovs= v1tovs*v2tov1 Program 4: Computes Zin, Zout, and gains using two port t-parameters. % two-port analysis in terms of t-parameters function [zin, zout] =twoportt(t, zL, zs)
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['analysis of terminated twoport using t-parameters'] t11= t(1,1); t12=t(1,2); t21=t(2,1); t22=t(2,2); zin= (t11*zL + t12)/(t21*zL + t22) zout= (t22*zs + t12)/(t21*zs + t11) v2tov1= zL/(t11*zL + t12) v1tovs= zin/(zin+zs) v2tovs= v2tov1*v1tovs Program 5: converts z-parameters to t-parameters %converting z to t paramters (same sormulas as %converting t to z parameters) function [t,t11,t12,t21,t22] = ztot(z) z11=z(1,1); z12=z(1,2); z21=z(2,1); z22=z(2,2); deltaz= z11*z22 - z12*z21; t11= z11/z21; t12= deltaz/z21; t21= 1/z21; t22= z22/z21; t= [ t11 t12; t21 t22]; S OLUTION 20.1. For network a, the z-parameters are by inspection Z a = R 1 + R 3 R 3 R 3 R 2 + R 3 Similarly, for network b, the z-parameters are the same: Z b = R 1 + R 3 R 3 R 3 R 2 + R 3 The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are Z new = Z a + Z b = 2 R 1 + R 3 R 3 R 3 R 2 + R 3
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S OLUTION 20.2. For networks a and b, the y-parameters are by inspection Y a = Y b = 3 - 2 - 2 4 S Hence, their z-parameters are the inverse of the y-parameter matrix: Z a = Z b = 1 8 4 2 2 3 The interconnection of networks a and b conforms to figure 20.2b, which is a series interconnection. Hence, the new overall z-parameters are Z new = Z a + Z b = 1 4 4 2 2 3 S OLUTION 20.3. For network a consisting of the single inductor, Z a = s s s s For network b, we have Z b = Y b [ ] - 1 = 1 2 1 1 2 4
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DeCarlo Ch20 Solutions - SOLUTIONS PROBLEMS CHAPTER 20...

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