Let's solve **part (a)** in **2 steps** explanations after each. --- ## (a) Find the total energy and powers of the signals through the time domain signals. ### **Step 1: Write the formulas and set up the integrals** - The **energy** of a signal \(x(t)\) is \(E = \int_{-\infty}^{\infty} |x(t)|^2 dt\). - The **power** of a signal \(x(t)\) is \(P = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt\). Let's apply these to each signal: #### 1. \(x(t) = e^{-a|t|}\) \[ E_x = \int_{-\infty}^{\infty} |e^{-a|t|}|^2 dt = \int_{-\infty}^{\infty} e^{-2a|t|} dt \] #### 2. \(y(t) = \cos(2\pi f t)e^{-a|t|}\) \[ E_y = \int_{-\infty}^{\infty} [\cos(2\pi f t)e^{-a|t|}]^2 dt = \int_{-\infty}^{\infty} \cos^2(2\pi f t) e^{-2a|t|} dt \] #### 3. \(z(t) = \cos(2\pi f t) + e^{-a|t|}\) \[ E_z = \int_{-\infty}^{\infty} \left[\cos(2\pi f t) + e^{-a|t|}\right]^2 dt \] **Explanation:** We've identified the correct formula for both energy and power, and set up the integrals to solve for each signal. --- ### **Step 2: Calculate the integrals and determine energies and powers** #### 1. \(x(t) = e^{-a|t|}\) \[ E_x = \int_{-\infty}^{\infty} e^{-2a|t|} dt = 2 \int_{}^{\infty} e^{-2a t} dt = 2 \left[ \frac{e^{-2a t}}{-2a} \right]_^\infty = 2\left( - \frac{1}{-2a}\right) = \frac{1}{a} \] For power, \[ P_x = \lim_{T\to\infty} \frac{1}{2T} E_x = \quad \text{(since energy is finite, power is zero)} \] #### 2. \(y(t) = \cos(2\pi f t)e^{-a|t|}\) Use \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\): \[ E_y = \int_{-\infty}^{\infty} \cos^2(2\pi f t) e^{-2a|t|} dt = \int_{-\infty}^{\infty} \frac{1}{2} e^{-2a|t|} dt + \frac{1}{2} \int_{-\infty}^{\infty} \cos(4\pi f t) e^{-2a|t|} dt \] The first term is \(\frac{1}{2} E_x = \frac{1}{2a}\). The second term is the Fourier transform of \(e^{-2a|t|}\) at frequency \(2f\): \[ \int_{-\infty}^{\infty} \cos(4\pi f t) e^{-2a|t|} dt = \frac{2a}{a^2 + (2\pi (2f))^2} \] So, \[ E_y = \frac{1}{2a} + \frac{a}{a^2 + (4\pi f)^2} \] #### 3. \(z(t) = \cos(2\pi f t) + e^{-a|t|}\) Expand the square: \[ E_z = \int_{-\infty}^{\infty} [\cos^2(2\pi f t) + 2\cos(2\pi f t)e^{-a|t|} + e^{-2a|t|}] dt \] - First term: \(\int \cos^2(2\pi f t) dt\) diverges (infinite). - Second term: \(\int_{-\infty}^{\infty} 2\cos(2\pi f t)e^{-a|t|} dt = 2 \cdot \frac{a}{a^2 + (2\pi f)^2}\) - Third term: \(E_x = \frac{1}{a}\) But since \(\cos^2(2\pi f t)\) integrates to infinity, **energy is infinite**. However, power for a pure cosine is nonzero (see part b). **Explanation:** Evaluated the integrals to find the total energy for each signal. For \(x(t)\) and \(y(t)\), the energies are finite, for \(z(t)\) it is infinite due to the cosine term. --- ## **Final Answer:** - \(x(t):\) **Energy = \(\frac{1}{a}\), Power = ** - \(y(t):\) **Energy = \(\frac{1}{2a} + \frac{a}{a^2 + (4\pi f)^2}\), Power = ** - \(z(t):\) **Energy = \(\infty\), Power = Nonzero (see part b)** Let me know if you'd like to proceed to part (b)!

# Total Energy and Power of Signals ## Part (a): Total Energy and Powers of the Signals ### Step 1: Formulas and Setup 1. **Energy** of a signal \(x(t)\): \[ E = \int_{-\infty}^{\infty} |x(t)|^2 dt \] 2. **Power** of a signal \(x(t)\): \[ P = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt \] ### Step 2: Calculate Integrals #### 1. For \(x(t) = e^{-a|t|}\): \[ E_x = \int_{-\infty}^{\infty} e^{-2a|t|} dt = 2 \int_{0}^{\infty} e^{-2at} dt = \frac{1}{a} \] - **Power**: \(P_x = 0\) (finite energy) #### 2. For \(y(t) = \cos(2\pi ft)e^{-a|t|}\): Using \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\): \[ E_y = \int_{-\infty}^{\infty} \cos^2(2\pi ft) e^{-2a|t|} dt \] - First term: \(\frac{1}{2}E_x = \frac{1}{2a}\) - Second term involves the Fourier transform: \[ \int_{-\infty}^{\infty} \cos(4\pi ft)e^{-2a|t|} dt = \frac{2a}{a^2 + (4\pi f)^2} \] Thus, \[ E_y = \frac{1}{2a} + \frac{a}{a^2 + (4\pi f)^2} \] #### 3. For \(z(t) = \cos(2\pi ft) + e^{-a|t|}\): \[ E_z = \int_{-\infty}^{\infty} \left[\cos(2\pi ft) + e^{-a|t|}\right]^2 dt \] - The first term diverges (infinite energy). - The second term: \(E_x = \frac{1}{a}\) - Thus, total energy \(E_z = \infty\). **Final Energies and Powers:** - \(E_x = \frac{1}{a}, P_x = 0\) - \(E_y = \frac{1}{2a} + \frac{a}{a^2 + (4\pi f)^2}, P_y = 0\) - \(E_z = \infty, P_z \neq 0\) --- ## Part (b): Comment on Energy and Power Signals - **Energy Signals**: Signals with finite energy and zero power. - **Examples**: \(x(t)\) and \(y(t)\). - **Power Signals**: Signals with finite power, possibly infinite energy. - **Example**: \(z(t)\). --- ## Part (c): Fourier Transforms of the Signals ### 1. \(x(t) = e^{-a|t|}\) - **Fourier Transform**: \[ X(f) = \frac{2a}{a^2 + (2\pi f)^2} \] ### 2. \(y(t) = \cos(2\pi ft)e^{-a|t|}\) - **Fourier Transform**: \[ Y(f) = \frac{a}{a^2 + (2\pi f)^2} + \frac{a}{a^2 + (2\pi (f-2f))^2} \] ### 3. \(z(t) = \cos(2\pi ft) + e^{-a|t|}\) - The Fourier Transform consists of two parts, with divergence in the cosine term. **Sketches**: - Sketch \(X(f)\) as a Lorentzian centered at \(f=0\). - Sketch \(Y(f)\) showing two peaks at \(f\) and \(-f\). - Sketch of \(Z(f)\) will show spikes at cosine frequencies. --- ## Part (d): Total Energies through Frequency Domain ### Parseval's Theorem Verification - Parseval's theorem states: \[ E_x = \int_{-\infty}^{\infty} |X(f)|^2 df \] ### Verification Steps: 1. Compute the integrals for \(X(f)\), \(Y(f)\), and \(Z(f)\). 2. Show that the energies calculated in the frequency domain match those from the time domain. --- ## Part (e): MATLAB Verification 1. **Calculate Energies**: - Use MATLAB to compute the time-domain energies of signals. - Calculate Fourier transforms. 2. **Sketching**: - Use MATLAB's plotting functions to sketch the signals and their Fourier transforms. 3. **Calculate Energies**: - Using Fourier transform representations, compute energies again and compare results. --- This structured analysis will guide you through understanding the signals and verifying results through various methods. If you need more detailed calculations or MATLAB code, let me know!