| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine the pole(s) of the given digital filter, we need to find the roots of the denominator of the transfer function of the filter in the Z-domain. The given equation is in the time domain and can be transformed to the Z-domain as follows: Given equation: $y(n) = \frac{1}{3} (x(n) + x(n-1) + x(n-2))$ In the Z-domain, this becomes: $Y(z) = \frac{1}{3} (X(z) + z^{-1}X(z) + z^{-2}X(z))$ Simplifying, we get: $Y(z) = \frac{1}{3} (1 + z^{-1} + z^{-2}) X(z)$ The transfer function, $H(z)$, is defined as the ratio of the output $Y(z)$ to the input $X(z)$: $H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} (1 + z^{-1} + z^{-2})$ To find the poles of the filter, we need to find the roots of the denominator of $H(z)$, which is $1 + z^{-1} + z^{-2}$. However, for the denominator to be in standard polynomial form, we need to multiply by $z^2$ to clear the negative powers of $z$: $z^2(1 + z^{-1} + z^{-2}) = z^2 + z + 1$ Now, we need to find the roots of $z^2 + z + 1 = 0$. We can use the quadratic formula to find these roots: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 1$, and $c = 1$. Plugging these values into the quadratic formula gives: $z = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2}$ Since $\sqrt{-3}$ is an imaginary number, the roots of the polynomial are complex: $z = \frac{-1 \pm \sqrt{3}i}{2}$ Therefore, the poles of the given digital filter are: $z_1 = \frac{-1 + \sqrt{3}i}{2}$ $z_2 = \frac{-1 - \sqrt{3}i}{2}$ |
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| deepseekr1 | To determine the pole(s) of the digital filter given by the equation \( y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2)) \), we start by finding the transfer function \( H(z) \). First, we take the Z-transform of both sides of the equation. The Z-transform of \( y(n) \) is \( Y(z) \), and the Z-transforms of \( x(n) \), \( x(n-1) \), and \( x(n-2) \) are \( X(z) \), \( z^{-1}X(z) \), and \( z^{-2}X(z) \) respectively. This gives us: \[ Y(z) = \frac{1}{3} \left( X(z) + z^{-1}X(z) + z^{-2}X(z) \right) \] Factoring out \( X(z) \), we get: \[ Y(z) = \frac{1}{3} X(z) \left( 1 + z^{-1} + z^{-2} \right) \] The transfer function \( H(z) \) is then: \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) \] To express this as a rational function, we multiply the numerator and denominator by \( z^2 \): \[ H(z) = \frac{1}{3} \left( \frac{z^2 + z + 1}{z^2} \right) = \frac{z^2 + z + 1}{3z^2} \] The denominator of the transfer function is \( 3z^2 \). Setting the denominator equal to zero to find the poles: \[ 3z^2 = 0 \implies z^2 = 0 \implies z = 0 \] Since the denominator is \( 3z^2 \), the pole at \( z = 0 \) has multiplicity 2. Therefore, the filter has two poles at the origin. \[ \boxed{0} \] (with multiplicity 2) |
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| gpto1 | To find the poles of the given digital filter, we start by expressing the filter's difference equation in terms of its transfer function in the \( z \)-domain. **Given difference equation:** \[ y(n) = \frac{1}{3} \left[ x(n) + x(n-1) + x(n-2) \right] \] where \( y(n) \) is the output and \( x(n) \) is the input at sample \( n \). **Step 1: Apply the \( z \)-Transform** Taking the \( z \)-transform of both sides of the equation, and using the time-shifting property of the \( z \)-transform (\( x(n-k) \leftrightarrow X(z) z^{-k} \)), we get: \[ Y(z) = \frac{1}{3} \left[ X(z) + X(z) z^{-1} + X(z) z^{-2} \right] \] **Step 2: Express the Transfer Function \( H(z) \)** The transfer function \( H(z) \) is defined as the ratio of the output to the input in the \( z \)-domain: \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) \] **Step 3: Rewrite \( H(z) \) with Positive Powers of \( z \)** To find the poles and zeros, we prefer polynomials with positive powers of \( z \). Multiply numerator and denominator by \( z^{2} \) to eliminate negative exponents: \[ H(z) = \frac{1}{3} \left( z^{2} + z + 1 \right) z^{-2} = \frac{z^{2} + z + 1}{3 z^{2}} \] **Step 4: Identify the Denominator \( D(z) \)** The denominator \( D(z) \) of the transfer function determines the poles: \[ D(z) = 3 z^{2} \] **Step 5: Find the Poles** Set \( D(z) = 0 \) and solve for \( z \): \[ 3 z^{2} = 0 \\ z^{2} = 0 \\ z = 0 \] Thus, there is a pole at \( z = 0 \). Since the equation is \( z^{2} = 0 \), this pole has multiplicity 2 (a double pole). **Conclusion:** The filter has a **double pole at \( z = 0 \)**. **Answer:** The filter has a double pole at **z = 0**; that is, both poles are at the origin in the z-plane. |
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| GPT-4o | To determine the poles of the given digital filter, we need to derive its transfer function \( H(z) \) and analyze its denominator. The given difference equation of the digital filter is: \[ y(n) = \frac{1}{3} \left( x(n) + x(n-1) + x(n-2) \right) \] First, we take the Z-transform of both sides of the equation. Recall that the Z-transform of a sequence \( x(n) \) is denoted by \( X(z) \), and the Z-transform properties for shifting in time are: \[ \mathcal{Z}\{x(n-k)\} = z^{-k} X(z) \] Applying the Z-transform to the given equation: \[ \mathcal{Z}\{ y(n) \} = \frac{1}{3} \left( \mathcal{Z}\{ x(n) \} + \mathcal{Z}\{ x(n-1) \} + \mathcal{Z}\{ x(n-2) \} \right) \] This yields: \[ Y(z) = \frac{1}{3} \left( X(z) + z^{-1} X(z) + z^{-2} X(z) \right) \] Factor out \( X(z) \): \[ Y(z) = \frac{1}{3} X(z) \left( 1 + z^{-1} + z^{-2} \right) \] The transfer function \( H(z) \) is defined as the ratio of the output \( Y(z) \) to the input \( X(z) \): \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) \] To express \( H(z) \) with positive powers of \( z \), multiply by \( z^2/z^2 \): \[ H(z) = \frac{1}{3} \frac{z^2 + z + 1}{z^2} \] Thus: \[ H(z) = \frac{1}{3} \left( z^{-2} + z^{-1} + 1 \right) \] Rewrite it as: \[ H(z) = \frac{z^2 + z + 1}{3z^2} \] The transfer function is: \[ H(z) = \frac{z^2 + z + 1}{3z^2} \] The poles of the filter are the values of \( z \) that make the denominator equal to zero. The denominator of \( H(z) \) is \( 3z^2 \), which is zero when: \[ 3z^2 = 0 \] Solving for \( z \): \[ z^2 = 0 \] \[ z = 0 \] Therefore, the pole of the filter is at: \[ z = 0 \] In summary, the given digital filter has a single pole at \( z = 0 \). |
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| llama3.1-70B-Instruct | The transfer function of the filter is given by $$H(z)=\frac{Y(z)}{X(z)}=\frac{1}{3}\left(1+z^{-1}+z^{-2}\right)$$ $$\Rightarrow 3Y(z)=X(z)+z^{-1}X(z)+z^{-2}X(z)$$ $$\Rightarrow 3Y(z)-X(z)-z^{-1}X(z)-z^{-2}X(z)=0$$ $$\Rightarrow \left(3-z^{-1}-z^{-2}\right)Y(z)=X(z)$$ $$\Rightarrow \left(3z^{2}-z-1\right)Y(z)=z^{2}X(z)$$ $$\Rightarrow H(z)=\frac{Y(z)}{X(z)}=\frac{z^{2}}{3z^{2}-z-1}$$ $$\Rightarrow H(z)=\frac{z^{2}}{\left(3z+1\right)\left(z-1\right)}$$ The poles of the filter are the roots of the characteristic equation $$\left(3z+1\right)\left(z-1\right)=0$$ $$\Rightarrow z=-\frac{1}{3},\,1$$ |
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| llama3.1-405B-Instruct | To find the pole(s) of the digital filter, we need to convert the given equation into a transfer function and then analyze its denominator. The given equation is: $$y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2))$$ We can rewrite this equation as: $$y(n) = \frac{1}{3}x(n) + \frac{1}{3}x(n-1) + \frac{1}{3}x(n-2)$$ Taking the z-transform of both sides, we get: $$Y(z) = \frac{1}{3}X(z) + \frac{1}{3}z^{-1}X(z) + \frac{1}{3}z^{-2}X(z)$$ Now, we can factor out $X(z)$ from the right-hand side: $$Y(z) = \frac{1}{3}(1 + z^{-1} + z^{-2})X(z)$$ The transfer function of the filter is: $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(1 + z^{-1} + z^{-2})$$ To find the pole(s), we need to find the values of $z$ that make the denominator of $H(z)$ equal to zero. In this case, the denominator is simply $3$, which is a constant and does not depend on $z$. Therefore, there are no poles in the transfer function. So, the solution is: There are no poles in the digital filter. |
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| DeepSeek-Coder-V2 | To determine the pole(s) of the given digital filter, we need to analyze the transfer function of the filter. The given difference equation is: \[ y(n) = \frac{1}{3} (x(n) + x(n-1) + x(n-2)) \] First, we rewrite the difference equation in a more standard form: \[ 3y(n) = x(n) + x(n-1) + x(n-2) \] To find the transfer function, we take the Z-transform of both sides of the equation. The Z-transform of a signal \(x(n)\) is denoted as \(X(z)\), and the Z-transform of a delayed signal \(x(n-k)\) is \(z^{-k}X(z)\). Applying the Z-transform to the difference equation, we get: \[ 3Y(z) = X(z) + z^{-1}X(z) + z^{-2}X(z) \] Factor out \(X(z)\) from the right-hand side: \[ 3Y(z) = (1 + z^{-1} + z^{-2})X(z) \] Now, divide both sides by \(X(z)\) to get the transfer function \(H(z)\): \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z^{-1} + z^{-2}}{3} \] To express \(H(z)\) in a more convenient form, multiply the numerator and the denominator by \(z^2\): \[ H(z) = \frac{z^2 + z + 1}{3z^2} \] The poles of the filter are the values of \(z\) that make the denominator of the transfer function equal to zero. Therefore, we need to solve the equation: \[ 3z^2 = 0 \] This simplifies to: \[ z^2 = 0 \] Thus, the solution is: \[ z = 0 \] Since \(z = 0\) is a double root, the filter has a single pole at \(z = 0\). Therefore, the pole(s) of the filter are: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To determine the pole(s) of the given digital filter, we need to find the roots of the denominator of the transfer function of the filter in the Z-domain. The given equation is in the time domain and can be transformed to the Z-domain as follows: Given equation: $y(n) = \frac{1}{3} (x(n) + x(n-1) + x(n-2))$ In the Z-domain, this becomes: $Y(z) = \frac{1}{3} (X(z) + z^{-1}X(z) + z^{-2}X(z))$ Simplifying, we get: $Y(z) = \frac{1}{3} (1 + z^{-1} + z^{-2}) X(z)$ The transfer function, $H(z)$, is defined as the ratio of the output $Y(z)$ to the input $X(z)$: $H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} (1 + z^{-1} + z^{-2})$ To find the poles of the filter, we need to find the roots of the denominator of $H(z)$, which is $1 + z^{-1} + z^{-2}$. However, for the denominator to be in standard polynomial form, we need to multiply by $z^2$ to clear the negative powers of $z$: $z^2(1 + z^{-1} + z^{-2}) = z^2 + z + 1$ Now, we need to find the roots of $z^2 + z + 1 = 0$. We can use the quadratic formula to find these roots: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 1$, and $c = 1$. Plugging these values into the quadratic formula gives: $z = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2}$ Since $\sqrt{-3}$ is an imaginary number, the roots of the polynomial are complex: $z = \frac{-1 \pm \sqrt{3}i}{2}$ Therefore, the poles of the given digital filter are: $z_1 = \frac{-1 + \sqrt{3}i}{2}$ $z_2 = \frac{-1 - \sqrt{3}i}{2}$ |
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| baichuan4 | To determine the poles of the filter described by the equation \[y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2))\] we need to analyze the system in the Z-domain. We will start by writing the Z-transform of the output signal \(Y(z)\) in terms of the Z-transform of the input signal \(X(z)\). Taking the Z-transform of both sides of the equation, we get: \[Y(z) = \frac{1}{3}(X(z) + z^{-1}X(z) + z^{-2}X(z))\] This can be simplified to: \[Y(z) = \frac{1}{3}X(z)(1 + z^{-1} + z^{-2})\] Now, we can factor out \(X(z)\) and write the transfer function \(H(z)\) of the filter: \[H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(1 + z^{-1} + z^{-2})\] To find the poles of the filter, we need to set the denominator of \(H(z)\) to zero and solve for \(z\). However, since this filter only has a numerator (it's a FIR filter), the poles are at infinity, and the filter is inherently stable. In conclusion, the filter described by the equation \(y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2))\) does not have any finite poles, and thus, it is a stable filter. |
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| doubao | To find the poles, we first need to find the transfer function. The transfer function H(z) is obtained by taking the Z-transform of the given equation. Let X(z) and Y(z) be the Z-transforms of x(n) and y(n) respectively. Taking the Z-transform of the equation gives: Y(z) = 1/3(X(z) + z^(-1)X(z) + z^(-2)X(z)) H(z) = Y(z)/X(z) = 1/3(1 + z^(-1) + z^(-2)) To find the poles, we set the denominator equal to zero: 1 + z^(-1) + z^(-2) = 0 Multiplying through by z^2 gives: z^2 + z + 1 = 0 Using the quadratic formula, the roots (poles) are: z = (-1 ± √(1 - 4*1*1)) / (2*1) = (-1 ± √(-3)) / 2 = (-1 ± i√3) / 2. So the poles of the filter are (-1 ± i√3) / 2. |
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| minimax | To determine the poles of the given digital filter, we need to find the roots of the denominator of the filter's transfer function in the z-domain. The transfer function H(z) of a filter is defined as the Z-transform of the filter's impulse response, which in this case is the equation given for y(n). The filter's difference equation is: $$ y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2)) $$ Taking the Z-transform of both sides of the equation, we get: $$ Y(z) = \frac{1}{3}(X(z) + z^{-1}X(z) + z^{-2}X(z)) $$ Where Y(z) is the Z-transform of y(n), and X(z) is the Z-transform of x(n). We can factor out X(z) since it is common to all terms: $$ Y(z) = \frac{1}{3}X(z)(1 + z^{-1} + z^{-2}) $$ Now, we can express the transfer function H(z) as the ratio of the output to the input in the z-domain: $$ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(1 + z^{-1} + z^{-2}) $$ To simplify this, we can substitute $z^{-1}$ with $1/z$: $$ H(z) = \frac{1}{3}(1 + \frac{1}{z} + \frac{1}{z^2}) $$ Multiplying through by $z^2$ to clear the fractions, we get: $$ H(z) = \frac{1}{3}(z^2 + z + 1) $$ Now, the poles of the filter are the values of z that make the denominator of H(z) equal to zero. Since there is no denominator in the simplified expression of H(z), we need to look at the original difference equation to find the poles. The poles are the roots of the equation: $$ 1 + z^{-1} + z^{-2} = 0 $$ Multiplying through by $z^2$ to clear the fractions, we get: $$ z^2 + z + 1 = 0 $$ This is a quadratic equation in z, and we can solve for z using the quadratic formula: $$ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where a = 1, b = 1, and c = 1. Plugging these values into the quadratic formula gives: $$ z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} $$ $$ z = \frac{-1 \pm \sqrt{1 - 4}}{2} $$ $$ z = \frac{-1 \pm \sqrt{-3}}{2} $$ $$ z = \frac{-1 \pm j\sqrt{3}}{2} $$ So the poles of the filter are complex conjugates located at: $$ z = \frac{-1 + j\sqrt{3}}{2} \quad \text{and} \quad z = \frac{-1 - j\sqrt{3}}{2} $$ These are the poles of the given digital filter. |
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| kimi | To determine the pole(s) of the digital filter, we first need to find the transfer function of the filter, which relates the output signal $y(n)$ to the input signal $x(n)$. The given equation is: $$y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2))$$ This can be rewritten in terms of the Z-transform, which is a tool used to analyze linear time-invariant systems in the frequency domain. The Z-transform of $y(n)$ is denoted as $Y(z)$, and the Z-transform of $x(n)$ is denoted as $X(z)$. The Z-transform of the given equation is: $$Y(z) = \frac{1}{3}(X(z) + z^{-1}X(z) + z^{-2}X(z))$$ Factor out $X(z)$: $$Y(z) = \frac{1}{3}X(z)(1 + z^{-1} + z^{-2})$$ Now, we can find the transfer function $H(z)$, which is the ratio of the output to the input in the Z-domain: $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(1 + z^{-1} + z^{-2})$$ The poles of the filter are the values of $z$ for which the denominator of the transfer function is zero (since the poles are in the denominator of the transfer function in the Z-domain). However, in this case, the transfer function is a finite impulse response (FIR) filter, which means it has no poles, as the denominator is always non-zero. For an FIR filter like this one, the poles are at $z = \infty$, which is not a finite value. This is because the impulse response of an FIR filter is finite and does not repeat, so it does not have any poles. In summary, the given digital filter is an FIR filter with no poles in the finite Z-plane. |
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| wenxinyiyan | 首先，我们需要将给定的差分方程转换为Z域中的传递函数。给定的差分方程是 $y(n)=\frac{1}{3}(x(n)+x(n-1)+x(n-2))$ 将方程两边都进行Z变换，我们得到 $Y(Z)=\frac{1}{3}(X(Z)+X(Z)Z^{-1}+X(Z)Z^{-2})$ 这可以简化为 $Y(Z)=\frac{1}{3}(1+Z^{-1}+Z^{-2})X(Z)$ 因此，传递函数H(Z)是 $H(Z)=\frac{Y(Z)}{X(Z)}=\frac{1}{3}(1+Z^{-1}+Z^{-2})$ 为了找到极点，我们需要找到使分母为零的Z的值。在这个例子中，分母是3，它不会为零。因此，这个滤波器没有极点。 |
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| zhipu | To determine the poles of the given digital filter, we first need to express the filter equation in terms of the Z-transform. The Z-transform is a mathematical tool used to convert discrete-time domain signals and systems into the z-domain, which simplifies the analysis of system properties such as poles and zeros. The given filter equation is: \[ y(n) = \frac{1}{3}(x(n) + x(n-1) + x(n-2)) \] ### Step 1: Take the Z-transform of both sides The Z-transform of a discrete-time signal \( x(n) \) is denoted as \( X(z) \), and similarly for \( y(n) \) as \( Y(z) \). Applying the Z-transform to both sides of the equation, we get: \[ Y(z) = \frac{1}{3}(X(z) + X(z)z^{-1} + X(z)z^{-2}) \] ### Step 2: Factor out \( X(z) \) \[ Y(z) = \frac{1}{3}X(z)(1 + z^{-1} + z^{-2}) \] ### Step 3: Express the transfer function \( H(z) \) The transfer function \( H(z) \) is the ratio of the output Z-transform to the input Z-transform: \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(1 + z^{-1} + z^{-2}) \] ### Step 4: Simplify the transfer function \[ H(z) = \frac{1}{3}(1 + \frac{1}{z} + \frac{1}{z^2}) \] \[ H(z) = \frac{1}{3} \left( \frac{z^2 + z + 1}{z^2} \right) \] \[ H(z) = \frac{z^2 + z + 1}{3z^2} \] ### Step 5: Identify the poles The poles of the filter are the values of \( z \) that make the denominator of \( H(z) \) equal to zero. In this case, the denominator is \( 3z^2 \). Setting the denominator to zero: \[ 3z^2 = 0 \] \[ z^2 = 0 \] \[ z = 0 \] ### Conclusion The filter has a pole at \( z = 0 \). Since \( z = 0 \) is a repeated root (it appears twice in the denominator), we have two poles at \( z = 0 \). Therefore, the poles of the filter are: \[ z = 0 \] (with multiplicity 2) This indicates that the filter is a type of low-pass filter, as poles at the origin are characteristic of such filters. |
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| gpto1 | To find the poles of the given digital filter, we start by expressing the filter's difference equation in terms of its transfer function in the \( z \)-domain. **Given difference equation:** \[ y(n) = \frac{1}{3} \left[ x(n) + x(n-1) + x(n-2) \right] \] where \( y(n) \) is the output and \( x(n) \) is the input at sample \( n \). **Step 1: Apply the \( z \)-Transform** Taking the \( z \)-transform of both sides of the equation, and using the time-shifting property of the \( z \)-transform (\( x(n-k) \leftrightarrow X(z) z^{-k} \)), we get: \[ Y(z) = \frac{1}{3} \left[ X(z) + X(z) z^{-1} + X(z) z^{-2} \right] \] **Step 2: Express the Transfer Function \( H(z) \)** The transfer function \( H(z) \) is defined as the ratio of the output to the input in the \( z \)-domain: \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) \] **Step 3: Rewrite \( H(z) \) with Positive Powers of \( z \)** To find the poles and zeros, we prefer polynomials with positive powers of \( z \). Multiply numerator and denominator by \( z^{2} \) to eliminate negative exponents: \[ H(z) = \frac{1}{3} \left( z^{2} + z + 1 \right) z^{-2} = \frac{z^{2} + z + 1}{3 z^{2}} \] **Step 4: Identify the Denominator \( D(z) \)** The denominator \( D(z) \) of the transfer function determines the poles: \[ D(z) = 3 z^{2} \] **Step 5: Find the Poles** Set \( D(z) = 0 \) and solve for \( z \): \[ 3 z^{2} = 0 \\ z^{2} = 0 \\ z = 0 \] Thus, there is a pole at \( z = 0 \). Since the equation is \( z^{2} = 0 \), this pole has multiplicity 2 (a double pole). **Conclusion:** The filter has a **double pole at \( z = 0 \)**. **Answer:** The filter has a double pole at **z = 0**; that is, both poles are at the origin in the z-plane. |
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| deepseekv3 | To determine the poles of the digital filter, we first need to analyze its transfer function. The given difference equation is: \[ y(n) = \frac{1}{3} \left( x(n) + x(n-1) + x(n-2) \right) \] ### Step 1: Take the Z-transform Apply the Z-transform to both sides of the equation. Recall that the Z-transform of \( x(n-k) \) is \( z^{-k} X(z) \), and the Z-transform of \( y(n) \) is \( Y(z) \). Thus: \[ Y(z) = \frac{1}{3} \left( X(z) + z^{-1} X(z) + z^{-2} X(z) \right) \] ### Step 2: Factor out \( X(z) \) Factor \( X(z) \) from the right-hand side: \[ Y(z) = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) X(z) \] ### Step 3: Compute the transfer function The transfer function \( H(z) \) is defined as the ratio of the output \( Y(z) \) to the input \( X(z) \): \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3} \left( 1 + z^{-1} + z^{-2} \right) \] ### Step 4: Rewrite in terms of \( z \) To find the poles, rewrite the transfer function in terms of \( z \) instead of \( z^{-1} \): \[ H(z) = \frac{1 + z^{-1} + z^{-2}}{3} = \frac{z^2 + z + 1}{3z^2} \] ### Step 5: Find the poles The poles of the filter are the values of \( z \) that make the denominator of the transfer function zero. For the transfer function: \[ H(z) = \frac{z^2 + z + 1}{3z^2}, \] the denominator is \( 3z^2 \). Setting the denominator to zero: \[ 3z^2 = 0 \implies z^2 = 0 \implies z = 0. \] Thus, the filter has a double pole at \( z = 0 \). ### Final Answer The pole(s) of the filter are at \( z = 0 \) (a double pole). |
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