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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
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tree | ab291cffc65280e58ac82470ba63fbcca7805165 /29/CH12/EX12.23.ii/exa12_23_ii.sce | |
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diff --git a/29/CH12/EX12.23.ii/exa12_23_ii.sce b/29/CH12/EX12.23.ii/exa12_23_ii.sce new file mode 100755 index 000000000..8413448ea --- /dev/null +++ b/29/CH12/EX12.23.ii/exa12_23_ii.sce @@ -0,0 +1,20 @@ +//caption:stability_using_Nyquist_criterion +//example 12_23_ii +//page 535 +disp("for K=1") +g=(0.1*(s+10)*(s+40))/(s*(s+1)*(s+4)); +g1=(0.1*(s1+10)*(s1+40))/(s1*(s1+1)*(s1+4)); +GH=syslin('c',g); +GH1=syslin('c',g1); +nyquist(GH); +nyquist(GH1); +//mtlb_axis([-3 0.5 -0.6 0.6]); +xtitle('Nyquist plot of (0.1*(s+10)*(s+40))/(s*(s+1)*(s+4))') +figure; +show_margins(GH,'nyquist') +disp("since the point(-1+%i0) is encircled twice clockwise by Nyquist plot ,so N=2 and P=0(given)") +N=-2;//no. of encirclement of -1+%i0 by G(s)H(s) plot anticlockwise +P=0;//no. of poles of G(s)H(s) with positive real part +Z=P-N;//np.of zeros of 1+G(s)H(s)=0 with positive real part +disp(Z,"Z=") +disp("as Z=2,there are two roots of closed loop characterstics eq having positive real part, hence system is unstable.") |