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I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.

Edit: The reference of the question is to book DSP : Principles , Algorithms and Applications by John Proakis and D Manolakis pdf of the book is here Page no 203 and 204. The two paragraphs after formula 3.6.4 explains the difference between zero input response and natural response

I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.

Edit: The reference of the question is to book DSP : Principles , Algorithms and Applications by John Proakis and D Manolakis pdf of the book is here Page no 203 and 204. The two paragraphs after formula 3.6.4 explains the difference between zero input response and natural response

Thank you Peter and Matt for your answers and comments.

I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.

I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.

Edit: The reference of the question is to book DSP : Principles , Algorithms and Applications by John Proakis and D Manolakis pdf of the book is here Page no 203 and 204. The two paragraphs after formula 3.6.4 explains the difference between zero input response and natural response

I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is  : it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.However if the z-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(Z)B(Z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then $$Y(z) = \sum_{k=1}^{\\N} A_k/(1-p_kz^{-1}) + \sum_{k=1}^{\\L} Q_k/(1-q_kz^{-1})$$ which gives $$y(n) = \sum_{k=1}^{\\N} A_k(p_k)^{n}u(n) + \sum_{k=1}^{\\L} Q_k(q_k)^{n}u(n)$$ where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. Its very confusing to grasp the difference between zero input and natural response.

I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.