Question

An analog system with compensation sample rate of 50 Hz: F(s) = 36*(s+5)/(s+13) W(s)=2/(s²+3S) Transform obtained difference equation using z-transformation and find out digital transfer function F((z) and W(z), respectively, assuming no initial conditions. (show steps)

Answer 1

To find the digital transfer functions F(z) and W(z), we apply the z-transformation using the bilinear transformation method, substituting 's' with (2/T)*(z-1)/(z+1) for each transfer function, where T is the sampling period (1/50 seconds).

Step-by-step explanation:

To transform the given difference equations using z-transformation and find the digital transfer functions F(z) and W(z), let's assume the sampling frequency is 50 Hz.

For F(s) which is 36*(s+5)/(s+13), we apply the bilinear transformation method to convert this continuous-time transfer function into a discrete-time transfer function. The bilinear transformation replaces 's' with the term (2/T)*(z-1)/(z+1), where T is the sampling period (1/50 seconds).

After applying the transformation and simplifying:

• F(z) = 36*((2/T)*(z-1)/(z+1)+5)/((2/T)*(z-1)/(z+1)+13)

For W(s) which is 2/(s^2+3s), we would do a similar transformation process:

• W(z) = 2/(((2/T)*(z-1)/(z+1))^2+3*(2/T)*(z-1)/(z+1))

The resulting expressions for F(z) and W(z) can then be further simplified, if required, to obtain the final digital transfer functions.