As it has proven in some paper and books the time and frequency resolution of Morlet wavelet is :
$\Delta t=\frac{f_c\sqrt{f_b}}{2f_i}$
$\Delta f=\frac{1}{2\pi f_c \sqrt{f_b}}$
I don't know how these relations have obtained, I have asked it in detail in here.you can answer that if you want.
but lets accept these relationship for now. there is some misunderstanding for me in here about the concept of resolution.
for this case, by increasing the term $f_c\sqrt{f_b}$ we have finer frequency resolution but poor time resolution. I am sure about finer frequency and poor time resolution because my practical results meet it. but if you look at $\Delta t$ and $\Delta f$ relationships you receive that increasing the term $f_c\sqrt{f_b}$ leads to decrease $\Delta f$ and increase the $\Delta t$.
does it mean that smaller value of resolution spell to have finer resolution and vice verca?
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$\begingroup$ In your last sentence: Smaller value of what resolution? Finer resolution of what? For each resolution, exactly what are you measuring or describing? $\endgroup$– hotpaw2Apr 11, 2014 at 22:01
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$\begingroup$ I mean does it mean that smaller value of frequency resolution ($\Delta f$)spell to have finer frequency resolution? and moreover does it mean that smaller value of time resolution ($\Delta t$)spell to have finer time resolution? @hotpaw2 $\endgroup$– SAHApr 12, 2014 at 4:42
2 Answers
Indeed, that's the Heisenberg Uncertainty Principle - you can't have both very good frequency and time resolution. You always have to sacrifice something. In case of Short Time Fourier Transform it's straightforward, but for wavelets are being 'squeezed', which is changing their frequency resolution. Figure below describes more than a thousand words:
EDIT: Because you are still trying to understand why it is like that, then please mind that by increasing the frequency of the wavelet ($f_c$), you should also shrink it in time (decrease $f_b$). For more info please refer for example to: Paul Addison - The Illustrated Wavelet Transform Handbook. You can find there a following graphs (Section 2.12):
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$\begingroup$ Thanks for you info, but I am well aware of that. but this is nothing to do with my question. please check the question again. @jojek $\endgroup$– SAHApr 11, 2014 at 21:13
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1$\begingroup$ If this has nothing to do with your question, you need to rephrase your question. $\endgroup$ Apr 12, 2014 at 0:50
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$\begingroup$ @MackTuesday The question is fine, I am just asking something about the concept of resolution. $\Delta f$ is said frequency resolution in many paper. and also they said increasing $fc \sqrt{f_b}$ leads to increase(finer) the frequency resolution. but if you look at $\Delta f$ relationship,you will see that $\Delta f$ is proportional to inverse of $fc \sqrt{f_b}$. this happens for $\Delta t$ as well. $\endgroup$– SAHApr 12, 2014 at 4:54
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$\begingroup$ There's a problem with the diagrams in this answer. The left hand tiling is correct for a linear frequency axis. The right hand diagram for the wavelet tiling should then refer to the same frequency axis, but it doesn't. It is only correct for a log-scale frequency axis, because the time-frequency area of the tiles is constant, the lower and wider tiles would have to be less high, namely by a factor of 2 (for this dyadic lattice) for each row. Therefore I think the diagrams can be quite misleading for a novice. $\endgroup$ Apr 12, 2014 at 8:25
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1$\begingroup$ @Electricman, yes, I think calling them uncertainties or just deviations is fine. You can even call them resolutions, but then make clear that smaller values indicate higher resolution! $\endgroup$ Apr 12, 2014 at 19:27
Increasing the numerical value of delta X usually means the same thing as decreasing the resolution of X, not increasing the resolution of X. However delta X is often called "resolution" in spite if this inverse slope relationship. But you have to be clear in your context about whether you are increasing the numerical value of delta X, or the quality of delta X (its resolution of X), which can have opposite meanings.
