Let's carefully analyze the problem: ### Given: - Input \( x[n] \) - Output \( y[n] \): \( y[n] = 10,\, 4,\, 10,\, 4,\, 10,\, 4,\, \ldots \) (repeats) - System impulse response \( h[n] = [a, b, c] \) (so, length 3 FIR filter) - The values of \( a, b, c \) are to be determined. ### Step 1: Express output in terms of input and impulse response For an FIR system (length 3): \[ y[n] = a \cdot x[n] + b \cdot x[n-1] + c \cdot x[n-2] \] ### Step 2: Guess the input based on the output pattern The output alternates between 10 and 4, so let's **try the simplest possible input**: Suppose the input is also a periodic sequence, say: \( x[n] = 1, 2, 1, 2, 1, 2, \ldots \) Let's test this hypothesis. ### Step 3: Write convolution equations for the first few points Let's write equations for \( y[0], y[1], y[2], y[3], \ldots \), using initial conditions (assume \( x[-1] = x[-2] = 0 \)): #### \( n = 0 \): \[ y[0] = a\,x[0] + b\,x[-1] + c\,x[-2] = a \cdot x[0] = a \cdot 1 = 10 \] So, \[ a = 10 \] #### \( n = 1 \): \[ y[1] = a\,x[1] + b\,x[0] + c\,x[-1] = a \cdot 2 + b \cdot 1 \] But \( y[1] = 4 \): \[ a \cdot 2 + b \cdot 1 = 4 \] We already have \( a = 10 \): \[ 10 \cdot 2 + b = 4 \implies 20 + b = 4 \implies b = 4 - 20 = -16 \] #### \( n = 2 \): \[ y[2] = a\,x[2] + b\,x[1] + c\,x[0] = a \cdot 1 + b \cdot 2 + c \cdot 1 \] \( y[2] = 10 \): \[ a \cdot 1 + b \cdot 2 + c \cdot 1 = 10 \] Plug in \( a = 10 \), \( b = -16 \): \[ 10 + (-16) \cdot 2 + c = 10 \implies 10 - 32 + c = 10 \implies c = 10 + 32 - 10 = 32 \] #### Let's check \( y[3] \): \[ y[3] = a\,x[3] + b\,x[2] + c\,x[1] = a\cdot 2 + b\cdot 1 + c\cdot 2 \] Plug in values: \[ y[3] = 10 \cdot 2 + (-16) \cdot 1 + 32 \cdot 2 = 20 - 16 + 64 = 68 \] But the given pattern is \( 10, 4, 10, 4, \ldots \) -- **this does not match**. ### Step 4: Try a different input Suppose \( x[n] = 1, 0, 1, 0, \ldots \) (alternating 1 and 0). Try \( n = 0 \): \( y[0] = a \cdot 1 = 10 \) \( n = 1 \): \( y[1] = a \cdot 0 + b \cdot 1 = b \cdot 1 = 4 \rightarrow b = 4 \) \( n = 2 \): \( y[2] = a \cdot 1 + b \cdot 0 + c \cdot 1 = a + c = 10 \) \( a = 10 \implies c = 0 \) Check \( y[3] \): \( y[3] = a \cdot 0 + b \cdot 1 + c \cdot 0 = b = 4 \) This matches the pattern: **10, 4, 10, 4, ...** ### Step 5: Write the final answer The impulse response is: \[ h[n] = [a, b, c] = [10, 4, 0] \] --- ## **Final Answer** \[ \boxed{ a = 10,\quad b = 4,\quad c = 0 } \] --- ### **How was this solved?** - Guessed a periodic input that could produce the given periodic output. - Used convolution equations to solve for \( a, b, c \). - Verified the pattern continues.

Let's analyze the problem step by step with detailed explanations: ### Given: - The system is characterized by an impulse response \( h[n] = [a, b, c] \). - Input \( x[n] \) produces output \( y[n] = 10, 4, 10, 4, 10, 4, \ldots \), which alternates between 10 and 4. - The system is a finite impulse response (FIR) filter of length 3. ### Step 1: Express the output in terms of the input and impulse response For an FIR filter: \[ y[n] = a \cdot x[n] + b \cdot x[n-1] + c \cdot x[n-2] \] Our goal is to determine \( a, b, c \). ### Step 2: Make an educated guess about the input \( x[n] \) Since the output is alternating between 10 and 4, a natural assumption is that \( x[n] \) is also a periodic sequence with a pattern that can produce such outputs. Let's try assuming \( x[n] \) is a simple sequence: \( x[n] = 1, 0, 1, 0, \ldots \). This is a common approach to simplify calculations. ### Step 3: Write convolution equations for initial samples Assuming initial inputs outside the known sequence are zero: \( x[-1] = x[-2] = 0 \). Now, write equations for the first few output samples: - For \( n=0 \): \[ y[0] = a \cdot x[0] + b \cdot x[-1] + c \cdot x[-2] = a \cdot 1 + 0 + 0 = a \] Given \( y[0] = 10 \), we get: \[ a = 10 \] - For \( n=1 \): \[ y[1] = a \cdot x[1] + b \cdot x[0] + c \cdot x[-1] = a \cdot 0 + b \cdot 1 + 0 = b \] Given \( y[1] = 4 \), we find: \[ b = 4 \] - For \( n=2 \): \[ y[2] = a \cdot x[2] + b \cdot x[1] + c \cdot x[0] = a \cdot 1 + b \cdot 0 + c \cdot 1 = a + c \] Given \( y[2] = 10 \), and knowing \( a = 10 \): \[ 10 + c = 10 \implies c = 0 \] ### Step 4: Verify the pattern with subsequent outputs Check \( y[3] \): \[ y[3] = a \cdot x[3] + b \cdot x[2] + c \cdot x[1] = a \cdot 0 + b \cdot 1 + c \cdot 0 = b = 4 \] which matches the pattern \(10, 4, 10, 4, \ldots \). Thus, the sequence and the calculated impulse response are consistent. ### **Final Answer:** \[ \boxed{ a = 10,\quad b = 4,\quad c = 0 } \] ### **Summary:** - Assumed the input pattern based on the output's periodicity. - Used initial conditions to solve for the impulse response coefficients. - The impulse response that produces the given output pattern with the assumed input is \( h[n] = [10, 4, 0] \). --- **Note:** The key insight was recognizing the periodicity and choosing an input that aligns with the output pattern, simplifying the calculation of the impulse response.

Let's analyze the problem step by step with detailed explanations: ### Given: - The system has an impulse response \( h[n] = [a, b, c] \). - The input \( x[n] \) produces an output \( y[n] \) that alternates between 10 and 4: \( y[n] = 10, 4, 10, 4, 10, 4, \ldots \). - The system is a finite impulse response (FIR) filter of length 3. ### Step 1: Express the output in terms of the input and impulse response For an FIR filter: \[ y[n] = a \cdot x[n] + b \cdot x[n-1] + c \cdot x[n-2] \] Our goal is to find the values of \( a, b, c \). ### Step 2: Make an educated guess about the input \( x[n] \) Since the output is alternating between 10 and 4, let's assume the input sequence \( x[n] \) is also periodic with a pattern that can produce such an output. A simple and common choice is to suppose \( x[n] \) is a sequence of zeros and ones, such as: \[ x[n] = 1, 0, 1, 0, \ldots \] This choice will simplify the analysis. ### Step 3: Write the convolution equations for initial points Assuming initial input values outside the known sequence are zero: \( x[-1] = 0 \), \( x[-2] = 0 \). Now, write equations for the first few output samples: - For \( n=0 \): \( y[0] = a \times x[0] + b \times x[-1] + c \times x[-2] \) Substituting \( x[0]=1 \), \( x[-1]=0 \), \( x[-2]=0 \): \[ y[0] = a \times 1 + b \times 0 + c \times 0 = a \] Given \( y[0] = 10 \), we find: \[ a = 10 \] - For \( n=1 \): \( y[1] = a \times x[1] + b \times x[0] + c \times x[-1] \) Substituting \( x[1]=0 \), \( x[0]=1 \), \( x[-1]=0 \): \[ y[1] = a \times 0 + b \times 1 + c \times 0 = b \] Given \( y[1] = 4 \), we find: \[ b = 4 \] - For \( n=2 \): \( y[2] = a \times x[2] + b \times x[1] + c \times x[0] \) Substituting \( x[2]=1 \), \( x[1]=0 \), \( x[0]=1 \): \[ y[2] = a \times 1 + b \times 0 + c \times 1 = a + c \] Given \( y[2] = 10 \), and knowing \( a=10 \): \[ 10 + c = 10 \implies c=0 \] ### Step 4: Verify the pattern with subsequent outputs Check \( y[3] \): \[ y[3] = a \times x[3] + b \times x[2] + c \times x[1] \] Substituting \( x[3]=0 \), \( x[2]=1 \), \( x[1]=0 \): \[ y[3] = a \times 0 + b \times 1 + c \times 0 = b = 4 \] This matches the expected alternating pattern. ### **Final answer:** \[ \boxed{ a=10,\quad b=4,\quad c=0 } \] ### **Summary:** - Assumed a periodic input sequence \( x[n]=1,0,1,0,\ldots \) to match the output pattern. - Used initial conditions to solve for \( a, b, c \): - \( a=10 \) from the first output. - \( b=4 \) from the second output. - \( c=0 \) from the third output. - Verified that the pattern holds for subsequent samples. This approach confirms that the impulse response that produces the given alternating output is \( h[n] = [10, 4, 0] \).