Let's solve **Problem 1** step by step: ## Problem 1: Delta Function Integrals The Dirac delta function, \(\delta(t-a)\), has the sifting property: \[ \int_{-\infty}^{\infty} f(t)\delta(t-a)dt = f(a) \] if \(a\) is within the integration bounds; otherwise, the integral is 0. --- ### (a) \(\int_{-3}^{3} (2t^2 + 5)\delta(t-2)dt\) - The delta function "picks out" the value at \(t=2\), which is within the limits. \[ \int_{-3}^{3} (2t^2 + 5)\delta(t-2)dt = 2(2)^2 + 5 = 8 + 5 = \boxed{13} \] --- ### (b) \(\int_{-\infty}^{\infty} e^{-j\omega t}\delta(3t-6)dt\) First, rewrite \(\delta(3t-6)\): \[ \delta(3t-6) = \frac{1}{3}\delta\left(t-2\right) \] So: \[ \int_{-\infty}^{\infty} e^{-j\omega t}\delta(3t-6)dt = \int_{-\infty}^{\infty} e^{-j\omega t} \frac{1}{3}\delta(t-2)dt = \frac{1}{3}e^{-j\omega \cdot 2} \] So the answer is: \[ \boxed{\frac{1}{3}e^{-j2\omega}} \] --- ### (c) \(\int_{0}^{4} \cos\left(\frac{\pi t}{2}\right)\delta(2t-3)dt\) Rewrite \(\delta(2t-3)\): \[ \delta(2t-3) = \frac{1}{2}\delta\left(t - \frac{3}{2}\right) \] Check if \(t = \frac{3}{2}\) is in the integration bounds (\(0 \leq t \leq 4\)): Yes. So: \[ \int_{0}^{4} \cos\left(\frac{\pi t}{2}\right)\delta(2t-3)dt = \frac{1}{2}\cos\left(\frac{\pi}{2} \cdot \frac{3}{2}\right) = \frac{1}{2}\cos\left(\frac{3\pi}{4}\right) = \frac{1}{2} \left(-\frac{1}{\sqrt{2}}\right) = \boxed{-\frac{1}{2\sqrt{2}}} \] --- ### (d) \(\int_{-1}^{2} (t^3 + 4t + 1)\delta(4-2t)dt\) Rewrite \(\delta(4-2t)\): \[ \delta(4-2t) = \frac{1}{2}\delta\left(t-2\right) \] Check if \(t=2\) is in the interval \([-1,2]\): Yes (endpoint is included). So: \[ \int_{-1}^{2} (t^3 + 4t + 1)\delta(4-2t)dt = \frac{1}{2}\left[(2)^3 + 4(2) + 1\right] = \frac{1}{2}(8 + 8 + 1) = \frac{1}{2}(17) = \boxed{\frac{17}{2}} \] --- ## **Summary of Answers for Problem 1** (a) \(\boxed{13}\) (b) \(\boxed{\frac{1}{3}e^{-j2\omega}}\) (c) \(\boxed{-\frac{1}{2\sqrt{2}}}\) (d) \(\boxed{\frac{17}{2}}\) --- If you'd like, I can continue with Problems 2 and 3!

Let's proceed step-by-step with **Problem 2: Signal Correlation**. --- ## Problem 2: Signal Correlation Given two signals over \(0 < t < 2\): \[ x_1(t) = e^{t} \] \[ x_2(t) = \sin(\pi t) \] --- ### (a) **Autocorrelation of \(x_1(t)\):** The autocorrelation function \( R_{x_1x_1}(\tau) \) is defined as: \[ R_{x_1x_1}(\tau) = \int_{0}^{2} x_1(t) x_1(t + \tau) dt \] But since the signals are only defined over \(0 < t < 2\), the limits of integration depend on \(\tau\). To simplify, assume \(\tau\) such that the shifted signals are within bounds. **However**, for simplicity, let's consider the autocorrelation at \(\tau = 0\): \[ R_{x_1x_1}(0) = \int_{0}^{2} [e^{t}]^2 dt = \int_{0}^{2} e^{2t} dt \] Calculating: \[ \int_{0}^{2} e^{2t} dt = \frac{1}{2} e^{2t} \bigg|_{0}^{2} = \frac{1}{2} (e^{4} - 1) \] **Final answer:** \[ \boxed{ R_{x_1x_1}(0) = \frac{e^{4} - 1}{2} } \] --- ### (b) **Cross-correlation between \(x_1(t)\) and \(x_2(t)\):** The cross-correlation function: \[ R_{x_1 x_2}(\tau) = \int_{0}^{2} x_1(t) x_2(t + \tau) dt \] Again, to evaluate, choose \(\tau = 0\) for simplicity: \[ R_{x_1 x_2}(0) = \int_{0}^{2} e^{t} \sin(\pi t) dt \] This integral can be computed using integration by parts or complex exponentials. --- ### Computing \(\int_{0}^{2} e^{t} \sin(\pi t) dt\): Recall that: \[ \int e^{a t} \sin(b t) dt = \frac{e^{a t}}{a^2 + b^2} (a \sin b t - b \cos b t) + C \] Here, \(a=1\), \(b=\pi\): \[ \int e^{t} \sin(\pi t) dt = \frac{e^{t}}{1 + \pi^2} ( \sin \pi t - \pi \cos \pi t ) \] Evaluate from \(0\) to \(2\): \[ R_{x_1 x_2}(0) = \left[ \frac{e^{t}}{1 + \pi^2} ( \sin \pi t - \pi \cos \pi t ) \right]_0^2 \] Calculate at \(t=2\): - \(\sin \pi \cdot 2 = \sin 2\pi = 0\) - \(\cos 2\pi = 1\) At \(t=2\): \[ \frac{e^{2}}{1 + \pi^2} ( 0 - \pi \cdot 1 ) = - \frac{\pi e^{2}}{1 + \pi^2} \] At \(t=0\): - \(\sin 0 = 0\) - \(\cos 0 = 1\) \[ \frac{e^{0}}{1 + \pi^2} ( 0 - \pi \cdot 1 ) = - \frac{\pi}{1 + \pi^2} \] Subtracting: \[ R_{x_1 x_2}(0) = \left( - \frac{\pi e^{2}}{1 + \pi^2} \right) - \left( - \frac{\pi}{1 + \pi^2} \right) = \frac{\pi}{1 + \pi^2} \left( 1 - e^{2} \right) \] **Final answer:** \[ \boxed{ R_{x_1 x_2}(0) = \frac{\pi (1 - e^{2})}{1 + \pi^2} } \] --- ## **Summary:** | Part | Expression | Result | |---------|----------------------------|--------------------------------------------------------------| | (a) Autocorrelation of \(x_1(t)\) at \(\tau=0\) | \(\int_0^2 e^{2t} dt\) | \(\displaystyle \frac{e^{4} - 1}{2}\) | | (b) Cross-correlation at \(\tau=0\) | \(\int_0^2 e^{t} \sin(\pi t) dt\) | \(\displaystyle \frac{\pi (1 - e^{2})}{1 + \pi^2}\) | --- If you'd like, I can also assist with **Problem 3** on orthogonal basis signals!

Let's proceed step-by-step with **Problem 3: Signals and Orthogonal Basis Functions**. --- ## Problem 3: Signals and Orthogonal Basis Functions **Given:** Two basis signals over \(0 \leq t \leq T\): \[ \phi_1(t) = 1, \quad \phi_2(t) = t \] **Question:** Determine whether these signals are orthogonal over the interval \([0, T]\). --- ### Step 1: Recall the orthogonality condition Two functions \(\phi_m(t)\) and \(\phi_n(t)\) are orthogonal over \([a, b]\) if: \[ \int_a^b \phi_m(t) \phi_n(t) dt = 0 \quad \text{for } m \neq n \] --- ### Step 2: Compute the inner product Calculate: \[ I = \int_0^T \phi_1(t) \phi_2(t) dt = \int_0^T 1 \times t \, dt \] \[ I = \int_0^T t \, dt = \frac{T^2}{2} \] - Since \(I = \frac{T^2}{2}\), **which is zero only when \(T=0\)** (a trivial case), the functions are **not orthogonal** over \([0, T]\) unless \(T=0\). ### **Conclusion:** \[ \boxed{ \text{The functions } \phi_1(t) = 1 \text{ and } \phi_2(t) = t \text{ are orthogonal over } [0, T] \textbf{ only if } T=0 } \] which is trivial and not meaningful for signal analysis. --- ### **Optional extension:** **What if** we normalize these functions? Let's find their norms: \[ \|\phi_1\|^2 = \int_0^T 1^2 dt = T \] \[ \|\phi_2\|^2 = \int_0^T t^2 dt = \frac{T^3}{3} \] --- ### **Summary:** - The inner product \( \int_0^T 1 \times t dt = T^2/2 \neq 0 \) unless \(T=0\). - Therefore, **the functions are not orthogonal** over any interval with positive length. --- ## **Final notes:** - For the basis functions \(\phi_1(t) = 1\) and \(\phi_2(t) = t\), they are **not orthogonal** over \([0, T]\). --- **If you'd like further exploration** into orthogonal functions (e.g., Legendre polynomials, Fourier basis), or other functions, just ask!

Certainly! Here are 50-word theoretical explanations for each problem: **Problem 1 (Delta Function Integrals):** The Dirac delta function acts as an impulse that "samples" the integrand at specific points where its argument zeroes. Integrals involving \(\delta(t - a)\) evaluate the integrand at \(t = a\), provided \(a\) lies within the integral limits, simplifying the calculation significantly. **Problem 2 (Signal Correlation):** Signal correlation measures the similarity between signals as a function of time shift. Autocorrelation assesses a signal's self-similarity, while cross-correlation compares two different signals. These are integral-based functions, often involving exponential or sinusoidal functions, providing insights into signal structure and mutual dependencies. **Problem 3 (Orthogonal Basis Signals):** Orthogonality of basis functions implies their inner product over a defined interval equals zero, indicating independence in the function space. Orthogonal bases facilitate efficient signal representation and analysis (e.g., Fourier series), where non-overlapping components simplify computations and promote minimal redundancy in signal decomposition.