Activity for MissMulan
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #287302 |
Andy hi. I was taught it in university yesterday and I didn't understand it. (more) |
— | over 1 year ago |
| Comment | Post #286836 |
@Elliot Alderson Editted. (more) |
— | over 1 year ago |
| Comment | Post #286784 |
@a concerned citizen but arent they biasing resistors part of the input impedance of the circuit ?Why are we neglecting them? (more) |
— | almost 2 years ago |
| Comment | Post #286765 |
Why do I need the IV curve?I have approximated the tunnel diode to behave like a resistor $$ R_{d} = \frac{V_{T}}{I_{s}}$$ from Vf to Vv and from 0 to Vp (more) |
— | almost 2 years ago |
| Comment | Post #286745 |
@#36396 Editted. (more) |
— | almost 2 years ago |
| Comment | Post #286738 |
@#52987 you are correct a BJT doesnt have a CE junction I used too literally the small signal analysis of the BJT. (more) |
— | almost 2 years ago |
| Comment | Post #286730 |
It is a exercise I made up to learn about filters. (more) |
— | almost 2 years ago |
| Comment | Post #286724 |
@a concerned citizen it is a EE question.Why does it matter why I want to ask the question? (more) |
— | almost 2 years ago |
| Comment | Post #286721 |
Ya I agree Q will be really high. (more) |
— | almost 2 years ago |
| Comment | Post #286724 |
I am designing it for fun. (more) |
— | almost 2 years ago |
| Comment | Post #286721 |
No I don't mean center frequency I mean bandwidth of the pass band filter,the range of frequencies at which the attenuation is higher than - 3dB (more) |
— | almost 2 years ago |
| Comment | Post #286700 |
@a concerned citizen I am just trying to explain to you what is inside my head. (more) |
— | almost 2 years ago |
| Comment | Post #286700 |
@LvW "the transfer function is defined for sinusoidal signals only!" why is that true?Why isnt the transfer function applied for a rectangular signal?Just s will be equal to something else instead of jω then. (more) |
— | almost 2 years ago |
| Comment | Post #286700 |
@a concerned citizen if s isnt any longer equal to jω maybe with the new input signal becomes 2jω the frequency at which we have a pole changes. (more) |
— | almost 2 years ago |
| Comment | Post #286700 |
For different input signals the frequency of the pole will have different values ,the pole as a pole exists for some specific values of L,C s but the frequency of the pole changes. (more) |
— | almost 2 years ago |
| Comment | Post #286689 |
Isn't the gain inversely proportional to frequency from the rolloff frequency of a filter? Why "well past"? (more) |
— | almost 2 years ago |
| Comment | Post #286687 |
In reality yes, but I am assuming that the slope db/Hz is constant.It is for educational purposes only. (more) |
— | almost 2 years ago |
| Comment | Post #286681 |
Olin I said "I have to design a low-pass filter" because I never thought we would end up at this place talking about why I am building this low-pass filter,I thought you would just help me with its design if you dont like it I can edit the question. (more) |
— | almost 2 years ago |
| Comment | Post #286680 |
Is the corner frequency equal to the cutoff frequency.And yes signals with frequency > cutoff frequency will be attenuated not completely go to 0. (more) |
— | almost 2 years ago |
| Comment | Post #286681 |
I just picked the requirements in random , I dont have to design anything , I am doing it for fun but I dont know how to solve this problem. (more) |
— | almost 2 years ago |
| Comment | Post #286681 |
It isnt homework.I was just curious on the design of such a filter so here I am. (more) |
— | almost 2 years ago |
| Comment | Post #285325 |
Star-Star and Delta-Delta topologies (more) |
— | over 2 years ago |
| Comment | Post #285242 |
Yes the question is theoretical I want to understand what bandwidth of the signal means. (more) |
— | over 2 years ago |
| Comment | Post #285242 |
@#53661 i don't know if it is needed thats why i am asking.I dont have a clue on what "bandwidth of the signal" means here. (more) |
— | over 2 years ago |
| Comment | Post #285171 |
Where should I ask it then? (more) |
— | over 2 years ago |
| Comment | Post #284818 |
I didnt go to university today and we had lab hours.In the laboratory we use a Protek IDM 505 and I didnt measure some circuits which are for homework so I am searching it online to see if I can find anything. (more) |
— | over 2 years ago |
| Comment | Post #284253 |
Yes i am interested in the capacitance between the windings.How can we make it bigger? (more) |
— | over 2 years ago |
| Comment | Post #284514 |
I will put some labels (more) |
— | over 2 years ago |
| Comment | Post #283942 |
"...how to make it because there are not genuine current sources in nature...".I disagree . A photodiode is a current source in photovoltaic mode. (more) |
— | over 2 years ago |
| Comment | Post #284219 |
@Olin Done!It has been fixed (more) |
— | over 2 years ago |
| Comment | Post #284219 |
@Olin i dont think it is possible in Multisim NOT to make any zig-zag . (more) |
— | over 2 years ago |
| Comment | Post #284076 |
@Carl please clarify?Isnt it well written,correct or well presented? (more) |
— | over 2 years ago |
| Comment | Post #284076 |
Because it is useful if someone wants to ask something relevant to the question. (more) |
— | over 2 years ago |
| Comment | Post #284074 |
Yes oops I will edit the image. (more) |
— | over 2 years ago |
| Comment | Post #284018 |
They are not homework questions and I make the images on my own , it takes plenty of time in order to contribute to this site so please respect that. (more) |
— | over 2 years ago |
| Comment | Post #283960 |
@Andy aka if you read the answer a little bit below that statement (which is true) i say we have voltages due to mutual inductance-> we can say that a current exitting the doted end of the inductor can be translated to a voltage drop from the dotted end to the undotted end.
(more) |
— | over 2 years ago |
| Comment | Post #283958 |
I mean how inductive coupling changes the total series inductance. (more) |
— | over 2 years ago |
| Comment | Post #283956 |
It isn't dimensionless im just doing the math. (more) |
— | over 2 years ago |
| Comment | Post #283956 |
I am using a LateX editor so I cant do what you want. (more) |
— | over 2 years ago |
| Comment | Post #283956 |
The (missing)units are obvious if you read the post. (more) |
— | over 2 years ago |
| Comment | Post #283956 |
I dont see any errors maybe you can enlighten me (more) |
— | over 2 years ago |
| Comment | Post #283926 |
The circuit isnt wrong but what you have suggested is a improvement of the current source/switch relationship. (more) |
— | over 2 years ago |
| Comment | Post #283936 |
1) The current source provides 0.1A to the circuit but this could well have been a typo or you may have not seen it well which is partly my fault because I didnt write it brightly enough 2)the time costant is L1/R so the t/T must have units of Ohms*seconds/Henries (more) |
— | over 2 years ago |
| Comment | Post #283936 |
Olin maybe you may want to delete this post. (more) |
— | over 2 years ago |
| Comment | Post #283926 |
What do you mean by designators exactly!? (more) |
— | over 2 years ago |
| Comment | Post #283904 |
Done check it out now. (more) |
— | over 2 years ago |
| Comment | Post #283904 |
You are correct I will edit (more) |
— | over 2 years ago |
| Comment | Post #283436 |
Suppose we have a transistor connected in series with a magnetic tunnel junction.1st case:the output voltage is low(the transistor is off) (0V)this represents the case of no charge state.2nd case:the transistor is on and the 2 layers of ferromagnetic materials which create the magnetic junction, hav... (more) |
— | over 2 years ago |
| Comment | Post #283436 |
@#53072iot if the charge state is 0, then we dont hav to worry about the spin state.Spintronics provide us with 3 available states unlike electronics which provides us with only 2 states. (more) |
— | over 2 years ago |
| Comment | Post #283331 |
So a DC blocking capacitor to not allow any current flowing from the DC source to the AC source? (more) |
— | over 2 years ago |