- #1
saltine
- 84
- 0
Hi this is a general question about z-transform and block diagram:
Suppose Y(z) is the output, U(z) is the input, and H(z) is the transfer function, then:
Y(z) = H(z) U(z).
Suppose we start with an open-loop transfer function G(s) under unity feedback, then to go from s to z, say we assume a zero-order-hold component in the open loop path. and obtain G(z) of that path. Then, we compute H(z) by H(z) = G(z) /( 1+G(z) ). Now, we transform the input from U(s) to U(z). We multiply H(z) by U(z) to get Y(z). Once we get Y(z), we could transform it back to Y(s) or to y(t).
Is this correct so far?
My question is, suppose you have two transfer functions in cascade A(s) and B(s), such that the combined transfer function is AB(s), we know that AB(z) does not equal to A(z)B(z), why could I multiply H(z) and U(z) to get Y(z)? Why am I allowed to multiply there?
Is there something wrong in what I said?
- Thanks
Suppose Y(z) is the output, U(z) is the input, and H(z) is the transfer function, then:
Y(z) = H(z) U(z).
Suppose we start with an open-loop transfer function G(s) under unity feedback, then to go from s to z, say we assume a zero-order-hold component in the open loop path. and obtain G(z) of that path. Then, we compute H(z) by H(z) = G(z) /( 1+G(z) ). Now, we transform the input from U(s) to U(z). We multiply H(z) by U(z) to get Y(z). Once we get Y(z), we could transform it back to Y(s) or to y(t).
Is this correct so far?
My question is, suppose you have two transfer functions in cascade A(s) and B(s), such that the combined transfer function is AB(s), we know that AB(z) does not equal to A(z)B(z), why could I multiply H(z) and U(z) to get Y(z)? Why am I allowed to multiply there?
Is there something wrong in what I said?
- Thanks
