Post Undeleted by nidhin
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What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $Z$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1). Provided, $\mathbf{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


 

considering your exampleFor causal and stable system, $$H(z) = 1+z^{-1} $$ withthe ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

Inverse of this $H(z)$ exists either for $|z|>1$ (causal) or for $|z|<1$ (anti-causal) not for both. So $Z$ inverse of $H'(z)$ does not exists for samemust extend to infinity and ROC should contain the unit circle. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $Z$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1). Provided, $\mathbf{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


 

considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

Inverse of this $H(z)$ exists either for $|z|>1$ (causal) or for $|z|<1$ (anti-causal) not for both. So $Z$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $Z$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1).

For causal and stable system, the ROC of $H'(z)$ must extend to infinity and ROC should contain the unit circle.

    Post Deleted by nidhin
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What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $\mathcal{Z}$$Z$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum by convolving with $\mathcal{Z}$if you you have inverse of $H'(z)$transfer function as expressed in (1). Provided, $\mathcal{Z}$$\mathbf{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

ButInverse of this $\mathcal{Z}$$H(z)$ exists either for $|z|>1$ (causal) or for $|z|<1$ (anti-causal) not for both. So $Z$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $\mathcal{Z}$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum by convolving with $\mathcal{Z}$ inverse of $H'(z)$ as expressed in (1). Provided, $\mathcal{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

But $\mathcal{Z}$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $Z$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1). Provided, $\mathbf{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

Inverse of this $H(z)$ exists either for $|z|>1$ (causal) or for $|z|<1$ (anti-causal) not for both. So $Z$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

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What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. In Z domainTaking $\mathcal{Z}$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you haveby convolving with $\mathcal{Z}$ inverse transfer functionof $H'(z)$ as expressed in (1). Provided, Z$\mathcal{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

But Z$\mathcal{Z}$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. In Z domain, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum if you you have inverse transfer function as expressed in (1). Provided, Z inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

But Z inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

What you want is $$x(t) = (x(t)\otimes h(t))\otimes h'(t)$$ where $\otimes$ denotes convolution. Taking $\mathcal{Z}$ transform, $$X(z) = X(z) \times H(z) \times H'(z)$$ or $$H'(z) = \frac{1}{H(z)} \tag{1}$$

So you can deconvolve a convolution sum by convolving with $\mathcal{Z}$ inverse of $H'(z)$ as expressed in (1). Provided, $\mathcal{Z}$ inverse of $\mathbf{H'(z)}$ exists for same ROC.


considering your example, $$H(z) = 1+z^{-1} $$ with ROC = entire $Z$ plane except at $z=0$. $$H'(z) = \frac{1}{1+z^{-1}}$$

But $\mathcal{Z}$ inverse of $H'(z)$ does not exists for same ROC. That is why you could not find any impulse response that if convolved with $[1\ 1]$ would result in $[1]$.

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