To compute the average and RMS (Root Mean Square) values of the periodic signals shown in the image, the following steps will be performed:

1. Signal Analysis (Left Signal)

Characteristics:

• It is periodic with a period $T = 6 \, \text{units}$.
• It alternates between linear segments (slopes) and flat sections over $t$.

Steps to Find Average and RMS:

1. Average Value: $X_{\text{avg}} = \frac{1}{T} \int_{0}^{T} x(t) \, dt$

2. RMS Value: $X_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} x^2(t) \, dt}$

Breakdown into Regions:

• The function will be split over the period into distinct time intervals, and integrals will be calculated accordingly.

2. Signal Analysis (Right Signal)

Characteristics:

• It is periodic with a period $T = 8 \, \text{units}$.
• It contains flat and sloping regions over each period.

Steps to Find Average and RMS:

The same formulae for average and RMS values apply, split into intervals:

• Flat Regions and Slopes will be treated separately using piecewise integration.

Let me calculate the detailed values for these signals. Hold on! sympy as sp

Define the time variables and periodic signals for analysis

t = sp.symbols('t')

---- Left Signal ----

Period is 6

Signal defined piecewise over one period (0 to 6)

x1 = sp.Piecewise( (2t, (t >= 0) & (t < 1)), # Region 1: 2t (2 - 2*(t - 1), (t >= 1) & (t < 3)), # Region 2: linear decrease (-2, (t >= 3) & (t < 4)), # Region 3: constant -2 (-2 + 2*(t - 4), (t >= 4) & (t < 5)), # Region 4: linear increase (2*(6 - t), (t >= 5) & (t < 6)) # Region 5: linear decrease )

---- Right Signal ----

Period is 8

Signal defined piecewise over one period (0 to 8)

x2 = sp.Piecewise( (2*t, (t >= 0) & (t < 2)), # Region 1: linear increase (2, (t >= 2) & (t < 3)), # Region 2: constant 2 (2 - (t - 3), (t >= 3) & (t < 4)), # Region 3: linear decrease (0, (t >= 4) & (t < 8)) # Region 4: constant 0 )

Average and RMS formula

def average_and_rms(signal, T, t): avg = (1 / T) * sp.integrate(signal, (t, 0, T)) rms = sp.sqrt((1 / T) * sp.integrate(signal**2, (t, 0, T))) return sp.simplify(avg), sp.simplify(rms)

Calculate for the left signal (T = 6)

avg1, rms1 = average_and_rms(x1, 6, t)

Calculate for the right signal (T = 8)

avg2, rms2 = average_and_rms(x2, 8, t)

avg1, rms1, avg2, rms2### Results

Left Signal (Period $T = 6$):

• Average Value: $X_{\text{avg}} = -\frac{1}{6} \approx -0.167$
• RMS Value: $X_{\text{rms}} = \frac{4}{3} \approx 1.333$

Right Signal (Period $T = 8$):

• Average Value: $X_{\text{avg}} = 0.9375$
• RMS Value: $X_{\text{rms}} \approx 1.458$

Would you like me to provide a step-by-step explanation of the calculations, or further analyze these signals?

Related Questions

1. How do the slopes of the signals influence their average and RMS values?
2. What is the significance of the RMS value in periodic signals?
3. Can we derive energy and power from the RMS value of these signals?
4. How would the calculations change if the signals were non-periodic?
5. How do average and RMS values differ for sinusoidal signals?

Tip

RMS values are especially useful in electrical engineering to measure the effective value of alternating currents or voltages!