Recent content by equalP

O, I found my mistake... I forgot the R... \int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2\\ =\frac{\pi}{2R}e^{-R} [e^{\frac{2Rt}{\pi}}]_0^\phi\\ =\frac{\pi}{2R}e^{-R}(e^{\frac{2R\phi}{\pi}}-1)\\ \leq\frac{\pi}{2R}e^{-R}(e^R-1)\\ =\frac{\pi}{2R}(1-e^{-R})...

\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2\\ =\frac{\pi}{2}e^{-R} [e^{\frac{2t}{\pi}}]_0^\phi|\\ =\frac{\pi}{2}e^{-R}(e^{\frac{2\phi}{\pi}}-1)\\ \leq\frac{\pi}{2}e^{-R}(e-1) \lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\...

I want to ask a few questions. Why |(Re^{it})^n|\leq R^n, |e^{-Re^{it}}|\leq |e^{-R\cos(t)}| are ≤ but not =? And I do like this: |e^{-R\text{cos}\theta}|\leq|e^{-R(1-\frac{2\theta}{\pi})}| \lim_{R\to\infty}|\int_{z=Re^{i\theta}} z^n e^{-z} dz|,\quad 0\leq \theta \leq \phi\\...

I found I typed something wrong (the inequality) and it should be the following: After plugging in and having \int_{\text{Im}z=0}z^ne^{-z}dz=2n! I got \int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_{ce^{i\phi}t}z^n e^{-z}dz-2n! with R\rightarrow \infty How can I use the inequality to get the bound for...

I get what you mean. But the path in ii does not at infinity. Or I can just change the R to infinity by having limit?

Homework Statement Use the principle of deformation of path to deduce \int_0^\infty t^n \textbf{cos}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{cos}((n+1)\phi) and \int_0^\infty t^n \textbf{sin}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{sin}((n+1)\phi) where a>0, b>0, c=\sqrt{a^2+b^2}, and...

As the tutorial has not covered this topic yet, I'm quite confused on this question. (there is another question about this topic too, but I would like to try that question after knowing how to do this question) Q.8 (a) (I'm not sure whether I will know how to do (b) after having (a) done, I...

I just get equation from the I-V characteristic graph and find that i_D=0.02v_D-0.01

Because I found the result is same as that I found in (a), I think it should be correct...

In (c), I first consider the dc source only and do the similar with (a) and found Vo=0.75V, Io=5mA In small signal model, r_d=\frac{1}{\frac{d i_d}{d v_d}}=\frac{1}{0.02}=50\Omega v_p=0.1\times \frac{100||50}{100+100||50}=0.025 i_p=0.02v_p-0.01=5\times 10^{-4} v_D=v_p...

Thank you very much and I have solved these 2 questions:smile:

Should I solve (b) with dc load line only? As the ac part of Vth is quite small... Vth=0.05cos(ωt)+1V Rth=50Ω Or I should draw a line with Vth=0.05+1V and a line with Vth=-0.05+1V?

This is going to be (b) now.

Oh, I found the problem. It should be i_D=0.02v_D-0.01 \frac{v_s+v_b-v_D}{100}=i_D+\frac{v_D}{100} 0.1cos(\omega t)+2-v_D=2v_D-1+v_D 0.1cos(\omega t)+3=4v_D v_D=0.025cos (\omega t)+0.75 i_D=20v_D-10=5\times 10^{-4} cos (\omega t)+5\times 10^{-3} Is it right? If yes, then how can I solve it...

Does it mean that I can only get the result with no cos(ωt) in (b) but not (a)? When I use the result in (a) to draw the line, I found that there may be something wrong in my calculation as the my calculated vd is in a small range near 0.5V... i_D=20v_D-10...