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I'm learning control theory and in my book it shows an example of this block diagram, the output c(z)=G2(Z)G1R(Z)/(1+G1G2H(z))

I'm quite confused because the last step I got is , I can't get the same answer after transfer this answer to Z domain.

Please help me enter image description here

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  • \$\begingroup\$ Show your steps. \$\endgroup\$
    – Chu
    Commented May 27, 2017 at 21:37
  • \$\begingroup\$ @Chu I just uploaded it, not sure if the work I had done is correct \$\endgroup\$ Commented May 27, 2017 at 22:14
  • \$\begingroup\$ Just take G1HC(z)G2(z) to the left side, and C(z) is a common factor. \$\endgroup\$
    – Chu
    Commented May 28, 2017 at 0:06
  • \$\begingroup\$ @Chu Can I do that? I thought G1HC(z) is a single function. The book says the Z transfer Z[G1(s)G2(s)]=G1G2(z) but I couldn't find the formula on conversion table I found... I just wondering what's the meaning of G1 standalone, is it just a constant? \$\endgroup\$ Commented May 28, 2017 at 0:16
  • \$\begingroup\$ Let the forward path (\$G_1\: sampler\: G_2\$) be \$G_3(s)\$, then obtain the loop equation in s, then do the z-transform. \$\endgroup\$
    – Chu
    Commented May 28, 2017 at 7:57

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Let \$\small G_3(s)=\left[ G_1(s)\rightarrow sampler \rightarrow G_2(s)\right]\$, and \$\small C(s)=\$ output signal of \$\small G_2(s)\$; then the loop equation is: $$\small C(s)= \frac{R(s)\:G_3(s)}{1+H(s)\:G_3(s)}$$

Taking z-transforms:

\$\small R(s)G_3(s)\rightarrow RG_3(z)\rightarrow RG_1(z)\:G_2(z)\$ (noting that \$\small RG_1\$ and \$\small G_2\$ may be separated due to the intervening sampler);

\$\small H(s)G_3(s)\rightarrow HG_1G_2(z)\$ (noting that \$\small HG_1\$ and \$\small G_2\$ cannot be separated since \$\small HG_1\$ is not a signal)

Hence: $$\small C(z)= \frac{RG_1(z)\:G_2(z)}{1+HG_1G_2(z)}$$

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  • \$\begingroup\$ What do you mean by "HG1" is not a signal? \$\endgroup\$
    – Roy
    Commented May 29, 2017 at 9:01
  • \$\begingroup\$ H and G1 are both transfer functions, and you can't sample a transfer function. RG1 is a signal, since it's the response of G1 to the input, R, so it can be sampled to produce RG1* \$\endgroup\$
    – Chu
    Commented May 29, 2017 at 9:37
  • \$\begingroup\$ Oh, I think I got it, the equation I derived seems wrong. \$\endgroup\$ Commented May 29, 2017 at 11:49

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