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For a pure sinusoid, \$\small v=Vcos(\omega t)\$ connected to a pure inductive load, \$\small L\$, we must have \$\small v=L\large \frac{di}{dt}\$, hence, integrating: $$\small i=\frac{1}{L}\int v\:d …

By inspection, adding a further \$\small 1 \:\Omega /2\: \Omega\$ stage always results in the same Thevenin resistance, but half the Thevenin voltage, of the preceding stage. So the output resistance …

The voltage, \$ v\$, across \$\small C\$ is equal to the the voltage across the \$\small R, L\$ series combination. The current in \$\small C\$ is \$i_1=\small C\large \frac{dv}{dt}\$ The current in …

For a series \$\small RLC\$ with the output taken across \$\small R\$, the amplitude ratio is: $$\small \lvert H(j\omega)\rvert=\frac{\omega RC}{\sqrt {(1-\omega^2LC)^2+\omega^2R^2C^2}} \:\:\:...\:(1 …

Apply the voltage divider: \$\frac{V_R}{V_S}=\frac{R}{R+j\omega L-\frac{j}{\omega C}}=\frac{\omega CR}{\omega CR+j\omega^2CL-j}=\frac{\omega CR}{\omega CR+j(\omega^2CL-1)}\$ Then take the magnitude.

Write A as the L1/C/L2 node voltage, and replace the components by their Laplace equivalent impedances: sL1, 1/sC, sL2, R (=RL) Node A: (A-V0)/sL2 + AsC +(A-Vi)/sL1 =0 ...(1) Node Vo: Vo/R + (Vo- …

The LT of the differential equation is obtained from the relationship \$\large\frac{dy(t)}{dt}\rightarrow \small sY(s)\$, giving: $$(s^2+3s+2)I_L(s)=\frac{2}{s^2}$$

Only graph 1 can be critically damped since there is an undershoot, but no subsequent overshoot. Assuming the capacitor has an initial condition, then the voltage across the three components in paral …

Use the Laplace transform. Several forms of the voltage across the parallel combination are possible, depending on component values. But this voltage is always zero at t=0 If the system is underdampe …

If \$\small R_1C_2\ll R_1C_1+R_2C_2\$ then the cut-off frequencies will be approximately \$\frac{1}{R_1C_1}\$ and \$\frac{1}{R_2C_2}\$ This is because the denominator of the TF (in Laplace form, sinc …

Generally, for an underdamped 2nd order system the 2% settling time is approximately \$\frac{4}{\zeta\:\omega_n}\$. But relating this ROT to, say, a RLC circuit is not intuitive since changing one com …

The number of states required for a full rank state space model is the same as the number of distinct energy storage elements (i.e. distinct inductors and capacitors). Typically, the states are the cu …

It's, arguably, easier to solve this in Laplace and, if the system were 1st order, the solution will be of the form: \$\small A(1-e^{-at})sin (\omega t+\phi)\$. You have an overdamped 2nd order system …

What you're seeing are the initial trajectories of exponential waveforms. If you reduce the square wave frequency, you'll get a better picture of what's happening. With the present frequency, the resp …

In using 1/s for the capacitor, you've forgotten about initial conditions. At t=0 the voltage across the capacitor is -10V. Either that voltage has been generated by V(s) (unlikely, if not impossible …