# Negetive Time

#### Rick

##### ॐ
Valued Senior Member
In Gain Vs Frequency graph of transistor,there's something terribly wrong.
the graph is as follows:
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the Frequency is X-axis and Gain(in DB) on Y-axis.now,the graph is constant fro a certain time that is to say a line parallel to X-axis and gain is constant,after a point there is a constant reduction of gain,and thus the slope becomes negetive.
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PROBLEM
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the slope of this graph as i analysed was nothing but time.now how is that possible?time remains 0 for a while and then becomes negetive when slope is negetive.that is to say T(final) is less than T(initial).this is ridiculous.am i wrong somewhere?or this graph is,but this graph is taken from a good book,viz Boylestad and Nashelsky.help me out please.
bye!

Hi zion,

I can't come up with the exact definition of the decibel right now (has been a long time since I used that unit ) but I remember it is definied in a logaritmic way. The slope of the graph has units:

dB/Hz = dB * s = 10*log(...)*s

If the slope is negative, you can just bring the minus sign inside the log-function, -log(x) = log(1/x). This fixes what you could interpret as time. But I don't entirely understand why you would want to involve time there ???

Bye!

Crisp

Crisp, Zion

For a transistor frequency vs gain graph, I just can't understand where time comes into it.
dB's are used to allow easy reading of the bandwidth coefficient, so you must be talking about a Gain reversal after a specific frequency OUTSIDE the 3dB point.
You can get -ve gain after the distortion frequency but to get -ve time you would need to have -ve frequency, because of the relationship between f,t and lamda, (which is not the same as 1/f).

Hi esp,

If you denote the graph of gain versus frequency by g(f) then the slope of that graph would be:

dg/df (derivative of g with respect to its variable, f)

In units, that derivative, or slope, would be: Gain / Frequency = dB / Hz = dB * s since Hz = 1/s. That's were time comes in, not directly on the graph, but when looking at the derivative... But I was also wondering why exactly time-evolution (since zion speaks of Tstart and Tfinal) is necessary in the discussion of those graphs.

Bye!

Crisp

what i mean to say is that although everything going on well in the graph,why is this slope coming out negetive,there must be something wrong somewhere.about t(initial) and final,i was saying that on a decreasing slope,if you take two points,where gain decreases,you find that t(finals)value is less than t(initial),so it seems to me that graph is theoritically wrong.because the first way to analyse a graph mathematically is its slope.

ok esp, about the time stuff, find the slope what will be the slope:

delta(gain)/delta(frequency).................(I)

so f=1/t;..................(1)
therefore: by putting the values of (1) in (I) we get,
slope = delta(time);
this means (t(final)-t(initial));
so isnt there an inaccuracy here or have got it wrong somewhere?

your inputs are welcome to us.

bye!

Crisp...STILL WAITING....

bye!

Hi Zion,

Patience is a virtue .... Sorry but I don't have the time anymore to read the board every day, so I just pop in every now and then.

There's an error in your reasoning; let's first introduce some symbols:

- Slope dg/df we denote by S
- Frequency f = 1/T where T is the period of the signal with frequency f
- g1, f1 = "initial" gain, frequency
- g2, f2 = "final" gain, frequency

Then your reasoning is (briefly summarized):
S = Delta(g) / Delta(f)

linear approximation of derivative:

S = (g2 - g1) / (f2 - f1)

S = (g2 - g1) * (T2 - T1)

This is negative so somehow there is a problem with time.

First of all, this reasoning is in error, because dividing by (f2 - f1) is not the same as multiplying by (T2 - T1) :

(f2 - f1)<sup>-1</sup> = (1/T2 - 1/T1))<sup>-1</sup> = T1*T2 / (T1 - T2)

Further remarks:

- If S is negative, this could very well be because (g2 - g1) is negative (eg. gain at frequency 2 is smaller than gain at frequency 1, which is often the case with amplifiers when you look at increasing frequency)

- The "time" dimension arises because the frequency (or period) term has a unit (s*s)/s = s. This only means that the result also has units of time, but not necessarily that the result says something about a time difference. The only "time difference" here is the term (T1 - T2) which says something about the periods associated with signals, but nothing about time evolution.

I hope this somehow clarifies the apparent paradox.

Bye!

Crisp

Got it.

Okay ,thanks Crisp.

i got it,now.

bye!