# Return to Answer

 Post Undeleted by nidhin occurred Jun 5 '14 at 8:39 5 deleted 359 characters in body edited Jun 5 '14 at 8:38 nidhin 42733 silver badges1212 bronze badges 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 occurred Jun 5 '14 at 7:20 4 added 122 characters in body edited Jun 5 '14 at 7:09 nidhin 42733 silver badges1212 bronze badges 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 $$[1]$$. 3 added 43 characters in body edited Jun 5 '14 at 6:17 nidhin 42733 silver badges1212 bronze badges 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 $$[1]$$. 2 added 318 characters in body edited Jun 5 '14 at 6:09 nidhin 42733 silver badges1212 bronze badges 1 answered Jun 5 '14 at 5:51 nidhin 42733 silver badges1212 bronze badges