The Concept Of Being A Gentleman in Charles Dickens Great Expectations :: Great Expectations Essays

'The beautiful young lady at Miss Havisham's, and she's more beautiful than anybody ever was and I admire her dreadfully and I want to be a gentleman on her account' - Pip (page 126) This is the turning point where Dickens advances the not so clear plot of the story. This is where Pip admits to Biddy he is in love with Estella and wants to become a gentleman. He is, at this point, doing it for the wrong reasons. He is doing it to impress Estella. When Pip is at Miss Havisham's he realises how much social classes actually matter. People who were orphans or had other jobs such as blacksmiths were regarded as people who could never become gentlemen. Estella makes it clear to Pip that he is in a lower social class than she is. 'Why, he is a common labouring boy' I thought I overheard Miss Havisham answer - only it seemed so unlikely - 'well? You can break his heart!' - Estella and Miss Havisham (page 57) The reason Miss Havisham says this is because she was hurt so badly by Compeyson she wants to hurt someone else as revenge for what happened to her. Pip at this point thinks a gentleman is someone who is rich and well respected; he does not think about the conduct of a gentleman. Pip thinks that people who are regarded as gentlemen are born into a decent family and have a socially respected occupation such as a lawyer or a clergyman. An example of what a true gentleman is would be Herbert. He sets the example to Pip and Pip looks up to him. Herbert's father knew exactly what Compeyson wanted when he tried to marry Miss Havisham. What Herbert says here was exactly what his father said. 'But that he was not to be, without ignorance or prejudice, mistaken for a gentleman because it is a principle of his that no man who was not a true gentleman at heart, ever was, since the world began, a true gentleman in manner'- Herbert (page 177) The Concept Of Being A Gentleman in Charles Dickens' Great Expectations :: Great Expectations Essays 'The beautiful young lady at Miss Havisham's, and she's more beautiful than anybody ever was and I admire her dreadfully and I want to be a gentleman on her account' - Pip (page 126) This is the turning point where Dickens advances the not so clear plot of the story. This is where Pip admits to Biddy he is in love with Estella and wants to become a gentleman. He is, at this point, doing it for the wrong reasons. He is doing it to impress Estella. When Pip is at Miss Havisham's he realises how much social classes actually matter. People who were orphans or had other jobs such as blacksmiths were regarded as people who could never become gentlemen. Estella makes it clear to Pip that he is in a lower social class than she is. 'Why, he is a common labouring boy' I thought I overheard Miss Havisham answer - only it seemed so unlikely - 'well? You can break his heart!' - Estella and Miss Havisham (page 57) The reason Miss Havisham says this is because she was hurt so badly by Compeyson she wants to hurt someone else as revenge for what happened to her. Pip at this point thinks a gentleman is someone who is rich and well respected; he does not think about the conduct of a gentleman. Pip thinks that people who are regarded as gentlemen are born into a decent family and have a socially respected occupation such as a lawyer or a clergyman. An example of what a true gentleman is would be Herbert. He sets the example to Pip and Pip looks up to him. Herbert's father knew exactly what Compeyson wanted when he tried to marry Miss Havisham. What Herbert says here was exactly what his father said. 'But that he was not to be, without ignorance or prejudice, mistaken for a gentleman because it is a principle of his that no man who was not a true gentleman at heart, ever was, since the world began, a true gentleman in manner'- Herbert (page 177)

Assignment Solution 01

North South University ETE 321 – Spring 2010 Instructor: Nahid Rahman Assignment #1 Total Marks: 100 Worth: 7. 5% 1. Consider the sinusoidally modulated DSB LC signal shown below. The carrier DSB-LC frequency is ? c and the message signal frequency is ? m. (a) Determine the modulation index m. Solution: Amax = 25 Amin = 5 ? 25 ? 5 = = 0. 67 + 25 + 5 (b) Write an expression for the modulated signal ? (t). Solution: 1 1 ) = (25 ? 5) = 10 = ( ? 2 2 1 1 ) = (25 + 5) = 15 = ( + 2 2 = + cos = cos + ( ) cos Assignment 1 Sol Page: 1 of 12 = 15 cos + 10 cos cos (c) Derive time domain expressions for the upper and lower sidebands.Solution: = 15 cos + 10 cos cos = 15 cos + 5 cos( + ) + 5 cos( ? ) Upper sideband: 5 cos( + ) ) Lower sideband: 5 cos( ? (d) Determine the total average power of the modulated signal , the carrier power and the two sidebands. Solution: Power of carrier signal = (15 cos )2 = + ? (15)2 2 cos cos = (cos( + ) + cos( ? )) 2 1 = 112. 5 W 2 (5)2 2 (5)2 Power of upper sideband = (5 cos( Power of lower sideband = (5 cos( ))2 = = 12. 5 W = 12. 5 W Power of modulated signal = 137. 5 W (e) Assuming that the message signal is a voltage signal, calculate the PEP (Peak Envelop Power) across a 100? load. Solution: PEP = 2 ))2 = eed to obtain the RMS value by dividing the peak by v2. (f) Determine the modulation efficiency ?. Solution: 12. 5 + 12. 5 = = 18. 18% 137. 5 Amax is the peak value of the modulated signal. To calculate the DC power, we = ( )2 v2 = (25 )2 v2 100 = 3. 125 W Assignment 1 Sol Page: 2 of 12 2. A DSB-SC modulated signal can be generated by multiplying the message signal with a periodic pulse generator and passing the resultant signal through a band-pass filter. = 2 cos 200 + cos 600 ( )= 1 2 + 2 ? (? 1) ? 1 cos ( =1 2 ? 1 (2 ? 1)) (a) Find the DSB-SC signal component in V(t). Solution: Input to the BPF: = ? 1 1 2 ? (? ) = ( ){ + cos ( (2 ? 1))} =1 2 ? 1 2 1 2 2 ? 1 2 1 = ( ) { + cos + †¢ cos 3 + †¢ cos 5 + other terms} 2 3 5 1 2 2 2 ( )+ ( ) cos ( ) cos 3 ( ) cos 5 = ? + + other terms 2 3 5 Output of the BPF: 2 = cos 2 = 2 cos 200 + cos 600 cos (b) Specify the unwanted components in V(t) that need to be removed by a BPF of suitable design. Solution: 1 2 2 ( ), ( ) cos 3 ( ) cos 5 , ,other terms 2 3 5 (c) Assume the carrier frequency is 500 Hz. Sketch the spectral density of the resulting DSB-SC waveform. Solution: = 2 cos 200 + cos 600 =2 ? 200 + 2 + 200 + ? 600 + + 600 Assignment 1 Sol Page: 3 of 12 = 2 = 1 cos ? = 2 1 = 2 ? 00 rad = 1000 rad See plot below. (d) In the sketch for Part (c), specify lower and upper sidebands. + + 1 2 + 1 2 = 1 + 1 Assignment 1 Sol Page: 4 of 12 3. Let f(t) be a real signal. The transmitter transmits the following modulated signal = cos + sin Where is the Hilbert transform of f(t). (a) Explain that the modulated signal is a lower sideband SSB signal using an example of = cos . Solution: Note that there was an error in the question. The frequency of f(t) should be ? m inst ead of ? c. Any students with a reasonable attempt to this question will be awarded full marks. However, the solution below refers to the corrected problem. cos = sin cos + sin sin = cos = cos ? Since, cos ? = cos cos + sin sin ? = + ? + ? + For ? > 0, the impulse function is located to the left of the carrier frequency. For ? < 0, the impulse function is located to the right of –? c. Therefore, the modulating function produces lower sideband signals. (b) Determine the frequency of the modulated signal. Solution: From the expression of , the frequency of the modulated signal is ? . Assignment 1 Sol Page: 5 of 12 4. An SSB signal is generated by modulating an fc = 1 MHz carrier by the message signal = 2 cos 2000? t + cos 4000? t . The amplitude of the carrier signal is Ac = 1. a) Determine the Hilbert transform of f(t). Solution: = 2 cos 2000? t + cos 4000? t ? ? = 2 cos 2000? t ? + cos 4000? t ? 2 2 = 2 sin 2000? t + sin 4000? t (b) Determine the time domain expression of the lower SSB and upper SSB signals. Solution: ?  ± t = cos ? sin ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t + cos 4k? t cos = 2 cos 2k? t + cos 4k? t cos ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t cos + cos 4k? t cos ? sin 4k? t sin ? 2 sin 2k? t sin = 2 cos 2k? t cos ? sin 2k? t sin + cos 4k? t cos ? sin 4k? t sin  ± 2000 + cos  ± 4000 = 2 cos (c) Sketch the magnitude spectrum of the lower SSB. Solution: ? t = 2 cos ? 2000 + cos ? 000 ? ? = 2 + ? 2000 + 2 ? + 2000 + + ? 4000 + ? + 4000 A 2? ? -? c -? c+4000? -? c+2000? ?c-4000? ?c-2000? ?c ? (d) The coherent detection of the lower SSB signal consists of multiplying the received modulated signal by cos followed by a low pass filter. If the local (receiver) oscillator generates a phase error ? (i. e. the message signal is now multiplied by cos + , write the expression at the output of the low-pass filter and discuss how the phase error will affect the demodulated signal. Solution: Assignment 1 Sol Page: 6 of 12 Input of the L PF: = = = cos cos cos + + cos cos A cos B = sin A cos B = = os + sin + + cos sin + cos + cos A + B + cos A ? B sin A + B + sin A ? B cos 2 + + sin ? + sin 2 + sin ? A = ? sin A = cos + cos 2 + ? sin + sin 2 + Output of the LPF: = = cos ? sin 2 cos 2000? t + cos 4000? t cos ? 1 2 sin 2000? t + sin 4000? t sin 2 = cos 2000? t cos + cos 4000? t cos 1 ? sin 2000? t sin ? sin 4000? t sin 2 1 cos 4000? t cos ? sin 4000? t sin 2 = cos 2000? t cos ? sin 2000? t sin + cos A cos B ? sin A sin B = cos A + B = cos 2000? t + ? + cos 4000? t + ? Assignment 1 Sol Page: 7 of 12 5. A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load.When a sinusoidal test tone with a peak amplitude of 5. 0 V is applied to the input of the modulator, it is found that the spectral line for each sideband in the magnitude spectrum for the output is 40% of the carrier line. Determine the following quantities in the output signal: (a) The modulation index. Solution: = + c os cos cos + cos cos = 1 1 = cos + cos ? + cos + 2 2 When a sinusoidal test tone with a peak amplitude of 5. 0 V is applied to the input of the modulator, it is found that the spectral line for each sideband in the magnitude spectrum for the output is 40% of the carrier line. 1 : = 0. 40 2 = : = 0. 0 (b) The peak amplitude of the lower sideband. Solution: A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load. = /v2 = 1000 Am is the amplitude (or â€Å"peak†) of the modulated signal. We need to use the rms value when calculating DC power. = 1000 2 = 10 = 316. 27 Peak amplitude of the sideband = = 158. 11 (c) The ratio of total sideband power to carrier power. Solution: Total power of the sidebands = = Carrier Power = Ratio = : = . cos : = cos . + = . 0. 8 ? . = = 0. 32 + cos + (d) The total power of output. Solution: Assignment 1 Sol Page: 8 of 12 Total Power = + = 33kW e) The total average power in the output if the peak amplit ude of the modulation sinusoid is reduced to 4. 0 V. Solution: Changing the modulation sinusoid peak amplitude will affect the modulation index. 4 = 5 4 = ? 0. 8 = 0. 64 5 Ratio of total sideband power to carrier power = . : = . 0. 64 = 0. 2048 Total Power = + = 30. 12kW Assignment 1 Sol Page: 9 of 12 6. Suppose that a message signal f(x) has bandwidth B Hz. If f(x) is modulated by one of the modulation schemes DSB-SC or SSB or VSB, then for demodulation, the receiver must generate a (local) carrier in phase and frequency synchronous with the incoming carrier. This is referred to as synchronous or coherent demodulation. ) (a) Draw a block diagram for the demodulator. Solution: (b) Assume that there is a frequency error in the local carrier (the phase is correctly estimated). Give the expression of the Fourier transform of the output of the demodulator for the case of DSB-SC modulation, sketch the spectrum of the output signal, and compare it with the spectrum of the original signal ( you may assume an arbitrary shape of F (w)). Solution: = cos ? cos +? = = = = = 1 2 ? = 2 ? ? ? ? ? cos cos ? cos ? os ? +? cos +? +? cos 2 +? + cos 2 + ? + ? 2 ? +? Without frequency error: Assignment 1 Sol Page: 10 of 12 With frequency error: (c) Repeat (b) for the case of SSB-SC modulation (you may do so by choosing either upper SSB or lower SSB). Solution: + sin = cos Input to the LPF: ? = cos = = = = ? ? ? ? +? + cos cos + ? cos cos cos cos ? 1 2 ? ? sin +? +? cos + + cos 2 ? +? sin sin +? +? cos cos +? +? + = 1 2 sin ? cos ? + +? + 1 2 ? ? sin 2 sin ? ? 2 ? = ? ? 2 ? ? ? 2 + + ? 2 ? +? Assignment 1 Sol Page: 11 of 12 With frequency error: d) Suppose that you are an engineer who responds to design a modulation system for a coarse environment in which it is difficult to generate a local carrier in frequency synchronous with the incoming carrier during some period of transmission. Which modulation system would you like to recommend, DSB-SC or SSB? Justify your answer. Solutio n: For DSB-SC, we notice a distortion in frequency. For SSB, we only observe a frequency shift. Therefore, it would be better to use SSB for a coarse environment. Assignment 1 Sol Page: 12 of 12

Hunger Games Survival Essay

The game is designed to kill you. From the first moment when you are dropped into the merciless jungle, you are actively trying to be killed. Forgetting the band of other survivors that are effortlessly searching for you, there are many terrors you must be ready for. To prepare for these, you must have a good shelter, a way to find food and water, and a positive mental attitude. A shelter is arguably the most important of the 10 essentials in survival. Without a shelter, you could get soaking in rain or snow, and subsequently, die from the moisture. You will get bitterly cold during the night and die in your sleep. So Katness knew she needed a good shelter. One that would keep her warm and dry, as well as mobile. She built two kinds of shelter; one was high in a tree, both to protect her from the indigenous, and to keep her off of the moist ground. She knew that sitting directly on cold or snow covered ground will increase the rate of heat loss drastically. So even in the tree, she put her jacket below her and her towel above her protect her from the elements, both below her and above her. The second shelter she built was a makeshift debris hut. She found the base of a huge redwoody type of tree, which provided a stable back and added protection from snowfall and climate protection. She, one again, added a blanket to the floor as well as many ribs on her debris hut and foliage both for camouflage and climate protection. Finding food and water in this Amazonian jungle proved difficult. As two of the ten essentials, food and water are necessary in survival. Luckily, she brought enough nutrient rich power bars with her to surpass the amount of time she needed to. She was also aware to not over work, and keep a 60% rule when searching so her power bars would be sufficient. Finding water, was much more difficult. Water is scarce in many jungle environments, so she had to take little bits of snow in a bottle and let it melt over time and drink it. Eating straight snow will actually make you thirstier than without it, good thing she new that. Even with a stable shelter, a substantial food and water supply, and a sense of safety, she was still finding it difficult to keep a positive mental attitude, or PMA. Without a positive mental attitude, someone in a survival situation could lose hope of being found or surviving the night, and could just die. That is why she does simple activities to keep a PMA. She thinks about here friends and family back home, her boyfriend who is also stuck in this jungle, and other activities to keep her motivated and keep her striving to survive this unfortunate event. Due to her knowledge of the 10 essentials, knowing how to build a shelter, and the power to keep a positive mental attitude, Katness made it out of this treacherous situation. This may have been a movie, but this is a real world situation that anyone could have found themselves in. The opportunity to learn about the 10 essentials, or how to build an adequate debris hut, is an opportunity that you do not want to miss. These life-saving tips have not only helped Katness make it out of the Hunger Games, but it may also help you make it out of a life or death situation of your own.