When I write A\text{rect}_{\Delta}(f-f_0) I mean it's a rect signal with amplitude A (usually a constant), duration Δ and is centered in f_0.
So in this case there are 2 rect signals, one centered in -4W on the left and the other one centered in +4W on the right. Both have a duration of 4W...
Hi everyone!
I was trying to solve an old exercise but in a different way. It asks to find the signal x(t) given the graphs of |X(f)| and arg[X(f)]. Please refer to the attachment.
My stategy this time is as follows:
1) find D(f) by deriving |X(f)|
2) antitransform D(f) and get d(t)
3)...
I thought of breaking the equation in 2 pieces:
\underbrace{\mathcal{F}^{-1}[ \frac{1}{16}f e^{j2\pi \frac{1}{144W}f }]}_{\text{1st addend}}+ \underbrace{\mathcal{F}^{-1}[\frac{3}{8}We^{j2\pi\frac{1}{144W}f } ]}_{\text{2nd addend}}
I know that \mathcal{F}[\delta(t-t_0) ]=e^{j2\pi t_0 f}...
Sorry I'm late in replying, I was little busy.
So X(f) should be:
X(f)=
\left\{
\begin{array}{ll}
(\frac{1}{16}f +\frac{3}{8}W)e^{-j(-\frac{2\pi}{144W}f)} &, \text{ for -6W $\leq$ f $\leq$ -2W} \\...
So you're saying the Amplitude should be something like:
\frac{1}{16}f +\frac{3}{8}W \mbox{, for -6W \leq f \leq -2W}
-\frac{1}{16}f +\frac{3}{8}W \mbox{, for 6W \leq f \leq 2W} \right.
As for the PHASE, it should be:
\phi(f)=-\frac{\pi}{72W}f
Which by the way means that at...
Hi everyone!
I'm not sure if I'm posting this question in the right section. Please don't be mad at me if I'm mistaken.
Can you please help me solve this problem?
Calculate the value of the signal x(t), given its spectrum (see figure in attachment), at the time t=2/W.
Attempted...
how? Where do I get another set of equations? seems as though the node method isn't needed to solve this problem. Do I have to make some adjustments and re-write these equations? If so, what are those adjustments?
\left[ \begin{array}{ c c c }
(\frac{1}{R} + sC) & 0 &...
Please see the pic I've attached. Did I get your suggestion right?
I thought that since the current is not flowing in the upper part of the circuit because of the open circuit represented by the capacitor, I could just eliminate it with a big red "X". This means that for t>0 the initial...
Ok. Z is the inverse matrix of Y:
Y= \left[ \begin{array}{cr}
2 & -1 \\
-1 & 2
\end{array} \right]
\rightarrow
Z=\left[ \begin{array}{cr}
\frac{2}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{2}{3}
\end{array} \right]
So this would be:
[Z][I]=[E]
\left[...
I'm not sure whether there are any initial conditions on the capacitor and the inductance.
After all Ig=1 for t ≤ 0. This certainly leads to some initial condition:
L has i(0^{-})\neq 0 and
C has v(0^{-})\neq 0
I'll go with node method. I'm attaching a new pic with the nodes I'm planning to consider. Before proceeding, could you pls check it out and tell me if it is a wise choice?
By the way there was a mistake in my previous post regarding the ideal transformer's equation: not I_3 but V_3 is equal to...
Given the following circuit: (see attached file)
find the Network Function between I_L, flowing as drawn in picture, AND I_g.
I guess all the calculation should lead up to an expression like this one:
I_L(s)=F(s) I_g(s), where F(s) is a function in "s" which corresponds to the NETWORK FUNCTION...
Let me try the second part. I'll refer to the 3rd diagram attached.
(2) I again need to know (Va-Vb) in order to find Q(fin). The current flows clock wise but I move counter clock wise because I have to go from A to B.
(V_A-V_B) +I(r+R) -f=0
(V_A-V_B)= f-I(r+R). (2.1)
In the above...
In the given circuit (1st diagram), the switch T is initially down (creates a connection). At a certain instant, the switch is turned off and for a while the circuit is kept in that state(final situation). Find the ΔQ=Q(fin)-Q(init) on the top plate of the capacitor while passing from the...
problem:
A capacitor has circle plates with radius R. the plates are at a distance h. between the plates there is a tension V. Find B, magnetic induction, at a distance r from the symmetry axis of the capacitor. See diagram attached.
what I did:
I know it has to be studied in two different...
find electrostatic force
problem:
there's a long straight wire with a λ1 charge. at some distance from it there's a bar charged with a λ2 charge. the bar is not parallel to the straight wire, instead it's inclined with an angle "alpha" between the dotted line, parallel to the wire, passing...
F=\int dF=\int E dq_2=\int_D^{D+a} \frac{\lambda}{2\pi\varepsilon_0x}\sigma b dx
is it like this?
When you say dq = \sigma dA = \sigma b dx
is it like you're dividing the layer into thin vertical, rectangular-shaped strips with height=b, and length=dx?
Could it instead be this?
(3.1) F=\int dF=\int E dq_2=\int \frac{\lambda}{2\pi\varepsilon_0D}dq_2=\frac{\lambda}{2\pi\varepsilon_0D}q_2=
q_2=\sigma(a\cdot b) \longrightarrow \ F=\frac{\lambda}{2\pi\varepsilon_0D} \sigma(a\cdot b)
if I'm not mistaken, the Electrical field at a distance D from the long straight wire should be this,
(2.1) E=\frac{\lambda}{2\pi\varepsilon_0D}
(2.2) Since dF=E\cdot dq_2 \text{ and } dq_2=\sigma ,
(2.3) F=\int dF=\int E dq_2=\int \frac{\lambda}{2\pi\varepsilon_0D}\sigma
ok? but...
[SOLVED] finding coulmb's force
Problem:
A very long straight wire has a uniform linear density charge, λ (every dx, linear unit, has a λ charge). At a distance D, from the wire, there's a rectangular layer with length=a and height=b and a uniform superficial density charge, σ (every dS...
Sorry I didn't get you.
The notation is
sign / exponent in Exc.2^{e-1} / mantissa norm. 1 & 2
1bit / e bits / (15-e) bits = Tot. 16 bits
example
suppose we find out e=6
and we want ot represent +1.0101 * 2^3 (doesn't matter what's its...
hey, I really REALLY, need a help with this exercise, got an exam day after tomorrow.
got an idea: suppose I do the 1 to 4 steps for each of the numbers? In the end I see which one "includes" the other and I choose that one.
Let's do it.
Let's take r, its Order of magnitude is 4...
Consider a floating point binary notation with 16 bits. From left to right, it consists of 1bit for the sign (0= "+"), e bits for the exponent represented in Excess~2^{e-1} and the remaining bits for the decimal part of the mantissa, normalized between 1 and 2 (1 \leq m <2).
a) Calculate the...
Why?
I agree with the 1st statement, but I'm still having doubts on the direction of the centrifugal force. Is it parallel to the ground (not the cone's surface) point out?
Could you just show it with a picture post? This will clear out all my doubts. Thank you.
m (\omega^2R) = N+G where N and G stand for Normal force and Gravity force.
omega² ∙R is the centripetal acceleration.
How do I fit in the friction force?
Is it right? I'm pretty much confused so please give me clear hints.
the exercise is a little difficult to explain in words, it would be much easier with a picture.
An iron cube is on the inside rough surface ( \mu_s=0.5 ) of a cone which is upside down. this cone rotates around its centre axis with a \omega=constant. between the axis and the surface there's...
Homework Statement
\omega=constant
R=0.2m
\theta=30°
\mu _s=0.5 between the iron cube and the surface
\omega_{max}=?
Homework Equations
Fx=Fsin(α) or cos(α)
Friction=µN
The Attempt at a Solution
in pink I drew the gravity force and the normal force.
Which other force...
Homework Statement
An object passes through point P0 in the instant t=0, going up with a velocity v0=3m/s along a tilted rough surface (dynamic friction coefficient µd=0.3) which forms an angle α=30° with the "ground". The length L of the slanted surface is practically endless.
Find
a)h...
I spoke to my teacher and I understood my mistake.
I was thinking that the reaction force to the surface was ALWAYS only "m∙g". Actually the normal force varies, it's not always "m∙g" or, better, it's not just that.
The gravity force balances the normal force AND the F_y=F∙sin30°
mg= N+...