If we replace s by jw in any equation so far.docVIP

/questionhtml If we replace s by jw in any equation so far ,we have the circuit’s response to an angular frequency . In the complex plot in Figure4 ,σ= 0 and hence s = j along the positive j axis .Thus ,the function’s value along this axis is the frequency response of the filter .We have sliced the function along the jw axis and emphasized the RC low-pass filter’s frequency-response curve by adding a heavy line for function values along the positive j axis . The more familar Bode plot looks different in form only because the frequency is expressed logarithmically . While the complex frequency’s imaginary part (jw) helps describe a response to AC signals, the real part (σ)helps describe a circuit’s transient response . Looking at Figure 3, we can therefore say something about the RC low-pass filter’s response as compared to that of the integrator .The low-pass filter’s transient response is more stable ,because its pole is in the negative-real half of the complex plane .That is ,the low-pass filter makes an infinte response . For low-pass filter , pole positions further down the –σaxis mean a higher w0 ,a shorter time constant ,and therefore a quicker transient response .Conversely ,a pole closer to the j axis cause a longer transient response . So far ,we have related the mathematical transfer functions of some simple circuits to thier associated poles and zeros in the complex-frequency plane . From these functions , we have derived the circuit’s frequency response (and hence its Bode plot)and also its transient response . Because both the integrator and the RC filter have only one s in the denominator of their transfer functions , they each have only one pole . That is, they are first-order filters . However , as we can see from Figure 2 , the first-order filter does not provide a very selective frequency response . To tailor a filter more closely to our needs , we must move on to higher orders. Form now on , we will describe the transfer functi